Chapter 7 Using Quadratic Functions to Model Data
Chapter 7 Using Quadratic Functions to Model Data
Chapter 7 Using Quadratic Functions to Model Data
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<strong>Chapter</strong> 7<br />
<strong>Using</strong> <strong>Quadratic</strong> <strong>Functions</strong> <strong>to</strong> <strong>Model</strong> <strong>Data</strong><br />
Homework 7.1<br />
1. 16 = 4⋅ 4 = 4 ⋅ 4 = 2⋅ 2 = 4<br />
3. 8 = 4⋅ 2 = 4 ⋅ 2 = 2 2<br />
5. 12 = 4⋅ 3 = 4 ⋅ 3 = 2 3<br />
7. 4 = 2 ⋅ 2 = 2 ⋅ 2 =<br />
2<br />
9 3 3 3 3 3<br />
9. 6 = 6 =<br />
6<br />
49 49 7<br />
11. 5 ⋅ 2 = 5 2 =<br />
5 2<br />
2 2 4 2<br />
13. 9 ⋅ 6 = 9 6 = 9 6 =<br />
3 6<br />
6 6 36 6 2<br />
27. x 2 = 32<br />
x = ±<br />
32<br />
x = ± 16⋅<br />
2<br />
x = ± 4 2<br />
29. x 2 = 300<br />
31.<br />
x = ±<br />
300<br />
x = ± 100⋅3<br />
x = ± 10 3<br />
2<br />
x − =<br />
48 0<br />
x<br />
33. x 2 = 9<br />
2<br />
= 48<br />
x = ±<br />
48<br />
x = ± 16⋅3<br />
x = ± 4 3<br />
; there are no real number solutions.<br />
15. 3 = 3 = 3 ⋅ 2 = 3 2 = 3 2 =<br />
3 2<br />
32 16⋅<br />
2 4 2 2 4 4 4⋅<br />
2 8<br />
35. x 2 + 25 = 0 means<br />
number solutions.<br />
2<br />
x = −<br />
25 ; there are no real<br />
17. 3 = 3 ⋅ 2 = 6 =<br />
6<br />
2 2 2 4 2<br />
19.<br />
21. x 2 = 36<br />
11 11 11 11 5 55 55<br />
= = = ⋅ = =<br />
20 20 4⋅5 2 5 5 2 25 10<br />
x = ±<br />
x = ± 6<br />
23. x 2 = 5<br />
x = ±<br />
36<br />
5<br />
25. x 2 − 3 = 0<br />
2<br />
x = 3<br />
x = ±<br />
3<br />
37. 4x 2 = 36<br />
x<br />
2<br />
= 9<br />
x = ±<br />
x = ± 3<br />
39. 5x 2 = 8<br />
x<br />
=<br />
5<br />
2 8<br />
9<br />
8 4⋅<br />
2 2 2<br />
x = ± = ± = ±<br />
5 5 5<br />
2 2 5 2 10<br />
x = ± ⋅ = ±<br />
5 5 25<br />
x = ±<br />
2 10<br />
5<br />
184
SSM: Intermediate Algebra Homework 7.1<br />
41. 3x 2 − 14 = 0<br />
51.<br />
2<br />
3(4 7) 9<br />
x +<br />
= −<br />
43.<br />
45.<br />
47.<br />
49.<br />
2<br />
3x<br />
14<br />
x<br />
2<br />
=<br />
=<br />
14<br />
3<br />
14 14<br />
x = ± = ±<br />
3 3<br />
14 3 42<br />
x = ± ⋅ = ±<br />
3 3 9<br />
x = ±<br />
42<br />
3<br />
2<br />
( x − 4) = 25<br />
x − 4 = ± 25<br />
x − 4 = ± 5<br />
x − 4 = 5 or x − 4 = −5<br />
x = 4 + 5 or x = 4 − 5<br />
x = 9 or x = −1<br />
2<br />
(8x + 3) = 36<br />
8x<br />
+ 3 = ± 36<br />
8x<br />
+ 3 = ± 6<br />
8x<br />
+ 3 = 6 or 8x<br />
+ 3 = −6<br />
8x<br />
= 3 or 8x<br />
= −9<br />
3 9<br />
x = or x = −<br />
8 8<br />
2<br />
(9x − 7) = 0<br />
9x<br />
− 7 = 0<br />
9x<br />
= 7<br />
x =<br />
7<br />
9<br />
2<br />
(8x − 3) + 6 = 90<br />
2<br />
(8x<br />
3) 84<br />
8x<br />
− 3 = ± 84<br />
8x<br />
− 3 = ± 4 ⋅21<br />
8x<br />
− 3 = ± 2 21<br />
8x<br />
= 3 ± 2 21<br />
x =<br />
− =<br />
3 ± 2 21<br />
8<br />
53.<br />
55.<br />
2<br />
(4 7) 3<br />
x +<br />
= − Since the square of a number is<br />
nonnegative, there are no real number<br />
solutions.<br />
2<br />
5( x − 6) + 3 = 23<br />
− =<br />
2<br />
5( x 6) 20<br />
− =<br />
2<br />
( x 6) 4<br />
x − 6 = ± 4<br />
x − 6 = ± 2<br />
x − 6 = 2 or x − 6 = −2<br />
x = 8 or x = 4<br />
− − + = −<br />
2<br />
3(4x<br />
9) 5 31<br />
− − = −<br />
2<br />
3(4x<br />
9) 36<br />
2<br />
(4x<br />
9) 12<br />
4x<br />
− 9 = ± 12<br />
4x<br />
− 9 = ± 4 ⋅3<br />
4x<br />
− 9 = ± 2 3<br />
4x<br />
= 9 ± 2 3<br />
x =<br />
− =<br />
9 ± 2 3<br />
4<br />
57. Solve for x when f(x) = 0<br />
2<br />
0 x 17<br />
x<br />
2<br />
= −<br />
= 17<br />
x = ±<br />
17<br />
− .<br />
The x-intercepts are ( 17,0 ) and ( 17,0)<br />
59. Solve for x when h(x) = 0<br />
2<br />
0 = ( x − 2) − 4<br />
− =<br />
2<br />
( x 2) 4<br />
x − 2 = ± 4<br />
x − 2 = ± 2<br />
x − 2 = 2 or x − 2 = −2<br />
x = 4 or x = 0<br />
The x-intercepts are ( 4,0 ) and ( )<br />
0,0 .<br />
185
Homework 7.1<br />
SSM: Intermediate Algebra<br />
61. Solve for x when f(x) = 0<br />
= − + +<br />
2<br />
0 3( x 1) 12<br />
2<br />
3( x 1) 12<br />
+ =<br />
+ =<br />
2<br />
( x 1) 4<br />
x + 1 = ± 4<br />
x + 1= ± 2<br />
x + 1 = 2 or x + 1 = −2<br />
x = 1 or x = −3<br />
The x-intercepts are ( 1,0 ) and ( 3,0)<br />
63. Solve for x when h(x) = 0<br />
= + +<br />
2<br />
0 2( x 5) 1<br />
2<br />
2( x 5) 1<br />
2<br />
+ = −<br />
( x 5)<br />
+ = −<br />
1<br />
2<br />
− .<br />
Since the square of a number is nonnegative, there<br />
are no real number solutions.<br />
65. Solve for x when f(x) = 0<br />
= − −<br />
2<br />
0 ( x 5) 9<br />
− =<br />
2<br />
( x 5) 9<br />
− = ±<br />
2<br />
( x 5) 9<br />
− = ±<br />
2<br />
( x 5) 3<br />
x − 5 = ± 3<br />
x − 5 = 3 or x − 5 = −3<br />
x = 8 or x = 2<br />
The x-intercepts are ( 8,0 ) and ( )<br />
2,0 . Since these<br />
points are symmetric, the average of the x-<br />
coordinates, 8 + 2 10 = = 5 , is the x-coordinate of<br />
2 2<br />
the vertex of f(x). Substitute 5 for x in the function<br />
<strong>to</strong> find the y-coordinate of the vertex:<br />
2<br />
f ( x ) = (5 − 5) − 9 = − 9<br />
Therefore, the vertex is (5, -9).<br />
67. Solve for x when h(x) = 0<br />
= − + +<br />
2<br />
0 2( x 7) 9<br />
2<br />
2( x 7) 9<br />
2<br />
+ =<br />
( x 7)<br />
+ =<br />
x + 7 = ±<br />
x + 7 = ±<br />
x + 7 = ±<br />
x = − 7 ±<br />
9<br />
2<br />
9<br />
2<br />
9<br />
2<br />
3<br />
2<br />
3<br />
2<br />
Rationalize the denomina<strong>to</strong>r:<br />
3 2 3 2<br />
⋅ =<br />
2 2 2<br />
3 2<br />
x = − 7 ±<br />
2<br />
x ≈ −4.88 or x ≈ −9.12<br />
So, the x-intercepts are ( − 4.88,0)<br />
and ( 9.12,0)<br />
− .<br />
Since these points are symmetric, the average of<br />
− 4.88 + ( −9.12)<br />
the x-coordinates,<br />
= − 7 , is the x-<br />
2<br />
coordinate of the vertex of f(x). Substitute -7 for x<br />
in the function <strong>to</strong> find the y-coordinate of the<br />
vertex:<br />
2<br />
h( x ) = −2( − 7 + 7) + 9 = 9<br />
Therefore, the vertex is (-7, 9).<br />
186
SSM: Intermediate Algebra Homework 7.2<br />
73.<br />
1<br />
2<br />
1<br />
2<br />
a ⎛ a ⎞ a a<br />
= ⎜ ⎟ = =<br />
1<br />
b ⎝ b ⎠<br />
2<br />
b<br />
b<br />
75. Answers may vary. Example:<br />
69. Solve for x when g(x) = 0<br />
2<br />
0 x 31<br />
x<br />
2<br />
= −<br />
= 31<br />
x = ±<br />
31<br />
x ≈ 5.57 or x ≈ −5.57<br />
So, the x-intercepts are ( − 31,0)<br />
and ( )<br />
31,0 .<br />
Since these points are symmetric, the average of<br />
31 + ( − 31)<br />
the x-coordinates,<br />
= 0 , is the x-<br />
2<br />
coordinate of the vertex of f(x). Substitute 0 for x in<br />
the function <strong>to</strong> find the y-coordinate of the vertex:<br />
2<br />
g( x ) = (0) − 31 = − 31<br />
You can solve quadratic equations by extracting<br />
the square root as long as they can be placed in<br />
2<br />
vertex form. (Samples: x − 81 = 0 or<br />
2<br />
2( x − 1) + 9 = 0 )<br />
You can solve quadratic equations by fac<strong>to</strong>ring as<br />
long as the quadratic expression contained in the<br />
equation is not prime. (Samples:<br />
2<br />
2<br />
x − 81 = ( x − 9)( x + 9) = 0 or x + 4x<br />
+ 4 = 0 ).<br />
Homework 7.2<br />
1.<br />
3.<br />
2<br />
⎛ 12 2<br />
⎜<br />
⎞ ⎟ = 6 = 36 =c<br />
⎝ 2 ⎠<br />
This expression is<br />
form is ( x + 6) 2<br />
.<br />
2<br />
⎛ −14<br />
⎞ 2<br />
⎜ ⎟ = ( − 7) = 49 = c<br />
⎝ 2 ⎠<br />
2<br />
x + 12x<br />
+ 36 and its fac<strong>to</strong>red<br />
Therefore, the vertex is (0, -31).<br />
This expression is<br />
form is ( x − 7) 2<br />
.<br />
2<br />
x − 14x<br />
+ 49 and its fac<strong>to</strong>red<br />
5.<br />
2 2<br />
⎛ −7 ⎞ ( −7) 49<br />
⎜ ⎟ = c<br />
2 = =<br />
⎝ 2 ⎠ 2 4<br />
This expression is<br />
⎛ 7 ⎞<br />
form is ⎜ x − ⎟<br />
⎝ 2 ⎠ .<br />
2<br />
2 49<br />
x − 7x<br />
+ and its fac<strong>to</strong>red<br />
4<br />
71. No, the student did not solve it correctly. They<br />
should have fac<strong>to</strong>red the left hand side first.<br />
2<br />
x − 10x<br />
+ 25 = 0<br />
( x − 5)( x − 5) = 0<br />
− =<br />
2<br />
( x 5) 0<br />
x − 5 = 0<br />
x = 5<br />
7.<br />
2 2<br />
⎛ 3 ⎞ 3 9<br />
⎜ ⎟ =<br />
2 = = c<br />
⎝ 2 ⎠ 2 4<br />
This expression is<br />
⎛ 3 ⎞<br />
form is ⎜ x + ⎟<br />
⎝ 2 ⎠ .<br />
2<br />
x + 3x<br />
+ and its fac<strong>to</strong>red<br />
4<br />
2 9<br />
187
Homework 7.2<br />
SSM: Intermediate Algebra<br />
9.<br />
11.<br />
2 2 2<br />
⎛ 1 1 ⎞ ⎛ 1 ⎞ 1 1<br />
⎜ ⋅ ⎟ = ⎜ ⎟ = = = c<br />
2<br />
⎝ 2 2 ⎠ ⎝ 4 ⎠ 4 16<br />
2 1 1<br />
This expression is x + x + and its fac<strong>to</strong>red<br />
2 16<br />
⎛ 1 ⎞<br />
form is ⎜ x + ⎟<br />
⎝ 4 ⎠ .<br />
2<br />
2 2 2 2<br />
⎛ −4 1 ⎞ ⎛ −4 ⎞ ⎛ −2 ⎞ ( −2) 4<br />
⎜ ⋅ ⎟ = ⎜ ⎟ = ⎜ ⎟ = = = c<br />
2<br />
⎝ 5 2 ⎠ ⎝ 10 ⎠ ⎝ 5 ⎠ 5 25<br />
2 4 4<br />
This expression is x − x + and its fac<strong>to</strong>red<br />
5 25<br />
⎛ 2 ⎞<br />
form is ⎜ x − ⎟<br />
⎝ 5 ⎠ .<br />
2<br />
2<br />
⎛ 6 ⎞ 2<br />
13. Since ⎜ ⎟ = 3 = 9 , add 9 <strong>to</strong> both sides of the<br />
⎝ 2 ⎠<br />
equation.<br />
2<br />
x + 6x<br />
= 1<br />
2<br />
x + 6x<br />
+ 9 = 1+<br />
9<br />
( x )<br />
2<br />
+ 3 = 10<br />
x + 3 = ± 10<br />
x = − 3 ± 10<br />
2<br />
⎛12<br />
⎞ 2<br />
15. Since ⎜ ⎟ = 6 = 36 , add 36 <strong>to</strong> both sides of the<br />
⎝ 2 ⎠<br />
equation.<br />
2<br />
x + 12x<br />
= 2<br />
2<br />
x + 12x<br />
+ 36 = 2 + 36<br />
( x )<br />
2<br />
+ 6 = 38<br />
x + 6 = ± 38<br />
x = − 6 ± 38<br />
2<br />
⎛ −2<br />
⎞ 2<br />
17. Since ⎜ ⎟ = ( − 1) = 1, add 1 <strong>to</strong> both sides of the<br />
⎝ 2 ⎠<br />
equation.<br />
2<br />
x − 2x<br />
= 10<br />
2<br />
x − 2x<br />
+ 1 = 10 + 1<br />
( x )<br />
2<br />
− 1 = 11<br />
x − 1 = ± 11<br />
x = 1±<br />
11<br />
19. First, change the equation <strong>to</strong> the form<br />
2 2<br />
x + bx = p; x − 18x<br />
= 3.<br />
2<br />
⎛ −18<br />
⎞ 2<br />
Since ⎜ ⎟ = ( − 9) = 81, add 81 <strong>to</strong> both sides of<br />
⎝ 2 ⎠<br />
the equation.<br />
2<br />
x − 18x<br />
= 3<br />
2<br />
x − 18x<br />
+ 81 = 3 + 81<br />
( x )<br />
2<br />
− 9 = 84<br />
x − 9 = ± 84<br />
x − 9 = ± 2 21<br />
x = 9 ± 2 21<br />
21. First, change the equation <strong>to</strong> the form<br />
2 2<br />
x + bx = p; x + 18x<br />
= 9.<br />
2<br />
⎛18<br />
⎞ 2<br />
Since ⎜ ⎟ = (9) = 81 , add 81 <strong>to</strong> both sides of<br />
⎝ 2 ⎠<br />
the equation.<br />
2<br />
x + 18x<br />
= 9<br />
2<br />
x − 18x<br />
+ 81 = 9 + 81<br />
( x )<br />
2<br />
+ 9 = 90<br />
x + 9 = ± 90<br />
x + 9 = ± 3 10<br />
x = − 9 ± 3 10<br />
23. First, change the equation <strong>to</strong> the form<br />
2 2<br />
x + bx = p; x + 4x<br />
= − 8.<br />
2<br />
⎛ 4 ⎞ 2<br />
Since ⎜ ⎟ = (2) = 4 , add 4 <strong>to</strong> both sides of the<br />
⎝ 2 ⎠<br />
equation.<br />
188
SSM: Intermediate Algebra Homework 7.2<br />
2<br />
x + 4x<br />
= −8<br />
2<br />
x + 4x<br />
+ 4 = − 8 + 4<br />
( x )<br />
2<br />
+ 2 = −4<br />
Since the square of a number is nonnegative, there<br />
are no real number solutions.<br />
2 2<br />
⎛ −7 ⎞ ( −7) 49<br />
25. Since ⎜ ⎟ =<br />
2 =<br />
⎝ 2 ⎠ 2 4<br />
of the equation.<br />
2<br />
x − 7x<br />
= 3<br />
2 49 49<br />
x − 7x<br />
+ = 3 +<br />
4 4<br />
2<br />
⎛ 7 ⎞ 12 49<br />
⎜ x − ⎟ = +<br />
⎝ 2 ⎠ 4 4<br />
⎛ 7 ⎞ 61<br />
⎜ x − ⎟ =<br />
⎝ 2 ⎠ 4<br />
2<br />
7 61<br />
x − = ±<br />
2 4<br />
7 61<br />
x − = ±<br />
2 2<br />
7 61<br />
x = ±<br />
2 2<br />
x =<br />
2 2<br />
7 ± 61<br />
2<br />
⎛ 5 ⎞ 5 25<br />
27. Since ⎜ ⎟ =<br />
2 =<br />
⎝ 2 ⎠ 2 4<br />
the equation.<br />
2<br />
x + 5x<br />
= 4<br />
2 25 25<br />
x + 5x<br />
+ = 4 +<br />
4 4<br />
2<br />
⎛ 5 ⎞ 16 25<br />
⎜ x + ⎟ = +<br />
⎝ 2 ⎠ 4 4<br />
⎛ 5 ⎞ 41<br />
⎜ x + =<br />
2<br />
⎟<br />
⎝ ⎠ 4<br />
2<br />
, add 25<br />
4<br />
, add 49<br />
4<br />
<strong>to</strong> both sides<br />
<strong>to</strong> both sides of<br />
5 41<br />
x + = ±<br />
2 4<br />
5 41<br />
x + = ±<br />
2 2<br />
5 41<br />
x = − ±<br />
2 2<br />
x =<br />
− 5 ± 41<br />
2<br />
29. First, change the equation <strong>to</strong> the form<br />
2 2<br />
x + bx = p; x + x = − 7.<br />
2 2<br />
⎛ 1 ⎞ 1 1<br />
Since ⎜ ⎟ =<br />
2 = , add 1 <strong>to</strong> both sides of the<br />
⎝ 2 ⎠ 2 4 4<br />
equation.<br />
2<br />
x + x = −7<br />
2 1 1<br />
x + x + = − 7 +<br />
4 4<br />
2<br />
⎛ 1 ⎞ −28 1<br />
⎜ x + ⎟ = +<br />
⎝ 2 ⎠ 4 4<br />
⎛ 1 ⎞ −27<br />
⎜ x + =<br />
2<br />
⎟<br />
⎝ ⎠ 4<br />
2<br />
Since the square of a number is nonnegative, there<br />
are no real number solutions.<br />
2 2 2<br />
⎛ −5 1 ⎞ ⎛ −5 ⎞ ( −5) 25<br />
31. Since ⎜ ⋅ ⎟ = ⎜ ⎟ = =<br />
2<br />
⎝ 2 2 ⎠ ⎝ 4 ⎠ 4 16<br />
both sides of the equation.<br />
2 5 1<br />
x − x =<br />
2 2<br />
2 5 25 1 25<br />
x − x + = +<br />
2 16 2 16<br />
2<br />
⎛ 5 ⎞ 8 25<br />
⎜ x − ⎟ = +<br />
⎝ 4 ⎠ 16 16<br />
⎛ 5 ⎞ 33<br />
⎜ x − ⎟ =<br />
⎝ 4 ⎠ 16<br />
2<br />
, add 25<br />
16 <strong>to</strong><br />
189
Homework 7.2<br />
SSM: Intermediate Algebra<br />
5 33<br />
x − = ±<br />
4 16<br />
5 33<br />
x − = ±<br />
4 4<br />
5 33<br />
x = ±<br />
4 4<br />
x =<br />
5 ± 33<br />
4<br />
33. Divide both sides by 2 so that a = 1:<br />
⎛ −4<br />
⎞ 2<br />
x − 4x<br />
= . Since ⎜ ⎟ = ( − 2)<br />
= 4 , add 4 <strong>to</strong><br />
2 ⎝ 2 ⎠<br />
both sides of the equation.<br />
2 3<br />
2 3<br />
x − 4x<br />
=<br />
2<br />
2 3<br />
x − 4x<br />
+ 4 = + 4<br />
2<br />
2 3 8<br />
( x − 2)<br />
= +<br />
2 2<br />
2 11<br />
( x − 2)<br />
=<br />
2<br />
x − 2 = ±<br />
x − 2 = ±<br />
11<br />
2<br />
11<br />
Rationalize the denomina<strong>to</strong>r:<br />
11 2 22<br />
⋅ =<br />
2 2 2<br />
x = 2 ±<br />
2<br />
22<br />
2<br />
4 22<br />
x = ±<br />
2 2<br />
4 ± 22<br />
x =<br />
2<br />
2<br />
35. Change the equation so that it is in the form<br />
2<br />
2<br />
ax + bx = p : 5x<br />
+ 15x<br />
= − 9 . Divide both sides<br />
by 5 so that a = 1:<br />
2 2<br />
⎛ 3 ⎞ 3 9<br />
⎜ ⎟ =<br />
2 = , add 9<br />
⎝ 2 ⎠ 2 4 4<br />
equation.<br />
2 9<br />
x + 3x<br />
= −<br />
5<br />
2 9 9 9<br />
x + 3x<br />
+ = − +<br />
4 5 4<br />
2<br />
⎛ 3 ⎞ 36 45<br />
⎜ x + ⎟ = − +<br />
⎝ 2 ⎠ 20 20<br />
⎛ 3 ⎞ 9<br />
⎜ x + ⎟ =<br />
⎝ 2 ⎠ 20<br />
2<br />
3 9<br />
x + = ±<br />
2 20<br />
3 9<br />
x + = ±<br />
2 20<br />
3 3<br />
x + = ±<br />
2 2 5<br />
x + 3x<br />
= − . Since<br />
5<br />
2 9<br />
Rationalize the denomina<strong>to</strong>r:<br />
3 5 3 5<br />
⋅ =<br />
2 5 5 10<br />
3 3 5<br />
x + = ±<br />
2 10<br />
3 3 5<br />
x = − ±<br />
2 10<br />
15 3 5<br />
x = − ±<br />
10 10<br />
− 15 ± 3 5<br />
x =<br />
10<br />
<strong>to</strong> both sides of the<br />
190
SSM: Intermediate Algebra Homework 7.2<br />
37. Divide both sides by 3 so that a = 1:<br />
2 2 2<br />
⎛ 4 1 ⎞ ⎛ 2 ⎞ 2 4<br />
Since ⎜ ⋅ ⎟ = ⎜ ⎟ = = , add 4 2<br />
⎝ 3 2 ⎠ ⎝ 3 ⎠ 3 9 9<br />
sides of the equation.<br />
2 4 5<br />
x + x =<br />
3 3<br />
2 4 4 5 4<br />
x + x + = +<br />
3 9 3 9<br />
2<br />
⎛ 2 ⎞ 15 4<br />
⎜ x + ⎟ = +<br />
⎝ 3 ⎠ 9 9<br />
⎛ 2 ⎞ 19<br />
⎜ x + ⎟ =<br />
⎝ 3 ⎠ 9<br />
2<br />
2 19<br />
x + = ±<br />
3 9<br />
2 19<br />
x + = ±<br />
3 9<br />
2 19<br />
x + = ±<br />
3 3<br />
2 19<br />
x = − ±<br />
3 3<br />
x =<br />
− 2 ± 19<br />
3<br />
2 4 5<br />
x + x = .<br />
3 3<br />
<strong>to</strong> both<br />
39. Change the equation so that it is in the form<br />
2<br />
2<br />
ax + bx = p : 2x<br />
− x = 7 . Divide both sides by 2<br />
so that a = 1:<br />
2 1 7<br />
x − x = . Since<br />
2 2<br />
2 2 2<br />
⎛ −1 1 ⎞ ⎛ −1 ⎞ ( −1) 1<br />
⎜ ⋅ ⎟ = ⎜ ⎟ = =<br />
⎝ ⎠ ⎝ ⎠<br />
sides of the equation.<br />
2<br />
2 2 4 4 16<br />
2 1 7<br />
x − x =<br />
2 2<br />
2 1 1 7 1<br />
x − x + = +<br />
2 16 2 16<br />
2<br />
⎛ 1 ⎞ 56 1<br />
⎜ x − ⎟ = +<br />
⎝ 4 ⎠ 16 16<br />
⎛ 1 ⎞ 57<br />
⎜ x − ⎟ =<br />
⎝ 4 ⎠ 16<br />
2<br />
, add 1<br />
16<br />
<strong>to</strong> both<br />
1 57<br />
x − = ±<br />
4 16<br />
1 57<br />
x − = ±<br />
4 16<br />
1 57<br />
x − = ±<br />
4 4<br />
1 57<br />
x = ±<br />
4 4<br />
x =<br />
1±<br />
57<br />
4<br />
41. Change the equation so that it is in the form<br />
2<br />
2<br />
ax + bx = p : 5x<br />
+ 7x<br />
= − 3 . Divide both sides by<br />
5 so that a = 1:<br />
2 2 2<br />
2 7 3<br />
x + x = − . Since<br />
5 5<br />
⎛ 7 1 ⎞ ⎛ 7 ⎞ 7 49<br />
⎜ ⋅ ⎟ = ⎜ ⎟ = =<br />
⎝ ⎠ ⎝ ⎠<br />
sides of the equation.<br />
2<br />
5 2 10 10 100<br />
2 7 3<br />
x + x = −<br />
5 5<br />
2 7 49 3 49<br />
x + x + = − +<br />
5 100 5 100<br />
2<br />
⎛ 7 ⎞ −60 49<br />
⎜ x + ⎟ = +<br />
⎝ 10 ⎠ 100 100<br />
⎛ 7 ⎞ −11<br />
⎜ x + ⎟ =<br />
⎝ 10 ⎠ 100<br />
2<br />
, add 49<br />
100<br />
<strong>to</strong> both<br />
Since the square of a number is nonnegative, there<br />
are no real number solutions.<br />
43. Divide both sides by 8 so that a = 1:<br />
2 2 2<br />
2 5 1<br />
x − x = .<br />
8 2<br />
⎛ −5 1 ⎞ ⎛ −5 ⎞ ( −5) 25<br />
Since ⎜ ⋅ ⎟ = ⎜ ⎟ = = , add<br />
2<br />
⎝ 8 2 ⎠ ⎝ 16 ⎠ 16 256<br />
<strong>to</strong> both sides of the equation.<br />
2 5 1<br />
x − x =<br />
8 2<br />
2 5 25 1 25<br />
x − x + = +<br />
8 256 2 256<br />
2<br />
⎛ 5 ⎞ 128 25<br />
⎜ x − ⎟ = +<br />
⎝ 16 ⎠ 256 256<br />
⎛ 5 ⎞ 153<br />
⎜ x − ⎟ =<br />
⎝ 16 ⎠ 256<br />
2<br />
25<br />
256<br />
191
Homework 7.2<br />
SSM: Intermediate Algebra<br />
5 153<br />
x − = ±<br />
16 256<br />
5 153<br />
x − = ±<br />
16 256<br />
5 9⋅17<br />
x − = ±<br />
16 16<br />
5 3 17<br />
x = ±<br />
16 16<br />
5 ± 3 17<br />
x =<br />
16<br />
45. First, change the equation so that it is in the form<br />
2<br />
ax + bx = p :<br />
2 x( x − 3) = 5 − x<br />
x − x = − x<br />
2<br />
2 6 5<br />
− =<br />
2<br />
2x<br />
5x<br />
5<br />
Divide both sides by 2, so that a = 1:<br />
2 5 5<br />
x − x =<br />
2 2<br />
2 2 2<br />
⎛ −5 1 ⎞ ⎛ −5 ⎞ ( −5) 25<br />
Since ⎜ ⋅ ⎟ = ⎜ ⎟ = =<br />
2<br />
⎝ 2 2 ⎠ ⎝ 4 ⎠ 4 16<br />
<strong>to</strong> both sides of the equation.<br />
2 5 5<br />
x − x =<br />
2 2<br />
2 5 25 5 25<br />
x − x + = +<br />
8 16 2 16<br />
2<br />
⎛ 5 ⎞ 40 25<br />
⎜ x − ⎟ = +<br />
⎝ 4 ⎠ 16 16<br />
⎛ 5 ⎞ 65<br />
⎜ x − ⎟ =<br />
⎝ 4 ⎠ 16<br />
2<br />
5 65<br />
x − = ±<br />
4 16<br />
5 65<br />
x − = ±<br />
4 16<br />
5 65<br />
x − = ±<br />
4 4<br />
5 65<br />
x = ±<br />
4 4<br />
x =<br />
5 ± 65<br />
4<br />
, add 25<br />
16<br />
47. First, change the equation so that it is in the form<br />
2<br />
ax + bx = p :<br />
x( x − 1) + 3 = 5( x + 1)<br />
− + 3 = 5 + 5<br />
2<br />
x x x<br />
2<br />
x − 6x<br />
= 2<br />
2<br />
⎛ −6<br />
⎞ 2<br />
= − =<br />
Since ( )<br />
⎜ ⎟<br />
⎝ 2 ⎠<br />
the equation.<br />
2<br />
x − 6x<br />
= 2<br />
2<br />
x − 6x<br />
+ 9 = 2 + 9<br />
( x )<br />
2<br />
− 3 = 11<br />
x − 3 = ± 11<br />
x = 3 ± 11<br />
3 9 , add 9 <strong>to</strong> both sides of<br />
49. No, the student did not solve the equation<br />
correctly. The student should have first divided<br />
both sides by 4 and then completed the square and<br />
extracted the roots. The correct solution is:<br />
+ =<br />
2<br />
4x<br />
6x<br />
1<br />
2 3 1<br />
x + x =<br />
2 4<br />
2 3 9 1 9<br />
x + x + = +<br />
2 16 4 16<br />
2<br />
⎛ 3 ⎞ 4 9<br />
⎜ x + ⎟ = +<br />
⎝ 4 ⎠ 16 16<br />
⎛ 3 ⎞ 13<br />
⎜ x + ⎟ =<br />
⎝ 4 ⎠ 16<br />
2<br />
3 13<br />
x + = ±<br />
4 16<br />
3 13<br />
x + = ±<br />
4 16<br />
3 13<br />
x + = ±<br />
4 4<br />
3 13<br />
x = − ±<br />
4 4<br />
− 3 ± 13<br />
x =<br />
4<br />
192
SSM: Intermediate Algebra Homework 7.3<br />
51. a.<br />
b.<br />
c.<br />
= + +<br />
2<br />
3 x 6x<br />
13<br />
2<br />
x + 6x<br />
= −10<br />
2<br />
x + 6x<br />
+ 9 = − 10 + 9<br />
( x )<br />
2<br />
+ 3 = −1<br />
Since the square of a number is nonnegative,<br />
there are no real number solutions.<br />
= + +<br />
2<br />
4 x 6x<br />
13<br />
2<br />
x + 6x<br />
+ 9 = 0<br />
( x )<br />
2<br />
+ 3 = 0<br />
x + 3 = 0<br />
x = −3<br />
= + +<br />
2<br />
5 x 6x<br />
13<br />
2<br />
x + 6x<br />
+ 8 = 0<br />
( x )( x )<br />
+ 2 + 4 = 0<br />
x + 2 = 0 or x + 4 = 0<br />
x = − 2 or x = −4<br />
= + +<br />
2<br />
0 x 10x<br />
25<br />
+ =<br />
2<br />
( x 5) 0<br />
x + 5 = 0<br />
x = −5<br />
59. Answers may vary.<br />
Homework 7.3<br />
1. a = 2, b = 5, c = − 2<br />
( )( )<br />
( )<br />
− ± − −<br />
x =<br />
2 2<br />
x =<br />
2<br />
5 5 4 2 2<br />
− 5 ± 41<br />
4<br />
3. a = 3, b = 7, c = − 2<br />
( )( )<br />
( )<br />
− ± − −<br />
x =<br />
2 3<br />
x =<br />
2<br />
7 7 4 3 2<br />
− 7 ± 73<br />
6<br />
53. Solve for x when f(x) = 0:<br />
= − +<br />
2<br />
0 x 8x<br />
3<br />
2<br />
x − 8x<br />
= −3<br />
2<br />
x − 8x<br />
+ 16 = − 3 + 16<br />
− =<br />
2<br />
( x 4) 13<br />
x − 4 = ± 13<br />
x = 4 ± 13<br />
55. Solve for x when f(x) = 0:<br />
= + +<br />
2<br />
0 x 4x<br />
5<br />
2<br />
x + 4x<br />
= −5<br />
2<br />
x + 4x<br />
+ 4 = − 5 + 4<br />
+ = −<br />
2<br />
( x 2) 1<br />
Since the square of a number is nonnegative there<br />
are no real number solutions. Therefore, there are<br />
no x-intercepts.<br />
57. Solve for x when f(x) = 0:<br />
5. a = 2, b = − 5, c = 1<br />
2<br />
( 5) ( 5) 4( 2)( 1)<br />
2( 2)<br />
− − ± − −<br />
x =<br />
x =<br />
5 ± 17<br />
4<br />
7. a = 3, b = − 2, c = 8<br />
2<br />
( 2) ( 2) 4( 3)( 8)<br />
2( 3)<br />
− − ± − −<br />
x =<br />
x =<br />
2 ± −92<br />
6<br />
Since the square root of a negative number is not a<br />
real number, there are no real solutions.<br />
2<br />
9. First write 2x<br />
+ 5x<br />
= 3 in the form<br />
2<br />
2<br />
ax + bx + c = 0 : 2x<br />
+ 5x<br />
− 3 = 0<br />
a = 2, b = 5, c = − 3<br />
193
Homework 7.3<br />
SSM: Intermediate Algebra<br />
( )( )<br />
( )<br />
2<br />
5 5 4 2 3<br />
− ± − −<br />
x =<br />
2 2<br />
− 5 ± 49<br />
x =<br />
4<br />
− 5 ± 7<br />
x =<br />
4<br />
− 5 + 7 −5 − 7<br />
x = or x =<br />
4 4<br />
2 −12<br />
x = or x =<br />
4 4<br />
x = 1 or x = − 3<br />
2<br />
11. a = 3, b = 0, c = − 17<br />
2<br />
( ) ( )( )<br />
2( 3)<br />
0 ± 0 − 4 3 −17<br />
x =<br />
x = ±<br />
x = ±<br />
x = ±<br />
204<br />
6<br />
2 51<br />
6<br />
51<br />
3<br />
2<br />
13. First write 2x<br />
= − 5x<br />
in the form<br />
2<br />
2x<br />
+ 5x<br />
= 0<br />
a = 2, b = 5, c = 0<br />
− ±<br />
x =<br />
− ( )( )<br />
( )<br />
2<br />
5 5 4 2 0<br />
2 2<br />
− 5 ± 25<br />
x =<br />
4<br />
− 5 ± 5<br />
x =<br />
4<br />
− 5 + 5 −5 − 5<br />
x = or x =<br />
4 4<br />
0 −10<br />
x = or x =<br />
4 4<br />
5<br />
x = 0 or x = −<br />
2<br />
2<br />
ax bx c<br />
+ + = 0 :<br />
2<br />
15. First write 4x<br />
= 2x<br />
+ 3 in the form<br />
2<br />
2<br />
ax + bx + c = 0 : 4x<br />
− 2x<br />
− 3 = 0<br />
a = 4, b = − 2, c = − 3<br />
( 2) (<br />
2<br />
2) 4( 4)( 3)<br />
2( 4)<br />
− − ± − − −<br />
x =<br />
2 ± 52<br />
x =<br />
8<br />
2 ± 2 13<br />
x =<br />
8<br />
( ± )<br />
2 1 13<br />
x =<br />
8<br />
1±<br />
13<br />
x =<br />
4<br />
17. First write x( x )<br />
2<br />
ax bx c<br />
+ + = 0 :<br />
( )<br />
− 3x x + 1 = 2<br />
2<br />
3x<br />
3x<br />
2<br />
− − =<br />
2<br />
3x<br />
3x<br />
2 0<br />
− − − =<br />
− 3 + 1 = 2 in the form<br />
a = − 3, b = − 3, c = − 2<br />
( 3) (<br />
2<br />
3) 4( 3)( 2)<br />
2( −3)<br />
− − ± − − − −<br />
x =<br />
3 ± −15<br />
x =<br />
−6<br />
Since the square root of a negative number is not a<br />
real number, there are no real solutions.<br />
2<br />
19. First write ( x )<br />
2<br />
ax bx c<br />
+ + = 0 :<br />
2<br />
( x )<br />
3 − 2 = 5x<br />
2<br />
3x<br />
6 5<br />
− =<br />
2<br />
3x<br />
5x<br />
6 0<br />
3 − 2 = 5x<br />
in the form<br />
x<br />
− − =<br />
a = 3, b = − 5, c = − 6<br />
194
SSM: Intermediate Algebra Homework 7.3<br />
( 5) (<br />
2<br />
5) 4( 3)( 6)<br />
2( 3)<br />
− − ± − − −<br />
x =<br />
5 ± 97<br />
x =<br />
6<br />
21. First write 3( x 4) 2x( x 1)<br />
2<br />
ax bx c<br />
+ + = 0 :<br />
( x + ) = − x( x − )<br />
3 4 2 1<br />
2<br />
3 12 2 2<br />
x + = − x + x<br />
2<br />
2x<br />
x 12 0<br />
+ + =<br />
a = 2, b = 1, c = 12<br />
− ±<br />
x =<br />
+ = − − in the form<br />
− ( )( )<br />
( )<br />
2<br />
1 1 4 2 12<br />
2 2<br />
− 1±<br />
−95<br />
x =<br />
4<br />
Since the square root of a negative number is not a<br />
real number, there are no real solutions.<br />
2<br />
23. First write 3 − 6x = x in the form<br />
2<br />
2<br />
ax + bx + c = 0 : x + 6x<br />
− 3 = 0<br />
a = 1, b = 6, c =− 3<br />
( − ± )<br />
( )( )<br />
( )<br />
2<br />
6 6 4 1 3<br />
− ± − −<br />
x =<br />
2 2<br />
x =<br />
− 6 ± 48<br />
2<br />
− ±<br />
x =<br />
6 4 3<br />
2<br />
2 3 3<br />
x =<br />
2<br />
x = − 3 ± 3<br />
25. a = 2, b = − 5, c = − 4<br />
( 5) (<br />
2<br />
5) 4( 2)( 4)<br />
2( 2)<br />
− − ± − − −<br />
x =<br />
5 ± 57<br />
x =<br />
4<br />
− 5 + 57 −5 − 57<br />
x = or x =<br />
4 4<br />
x ≈ 3.14 or x ≈ −0.64<br />
2<br />
27. First write 2.8x<br />
− 7.1x<br />
= 4.4 in the form<br />
2<br />
2<br />
ax + bx + c = 0 : 2.8x<br />
− 7.1x<br />
− 4.4 = 0<br />
a = 2.8, b = − 7.1, c = −4.4<br />
( 7.1) (<br />
2<br />
7.1) 4( 2.8)( 4.4)<br />
2( 2.8)<br />
− − ± − − −<br />
x =<br />
7.1±<br />
99.69<br />
x =<br />
5.6<br />
7.1+ 99.69 7.1−<br />
99.69<br />
x = or x =<br />
5.6 5.6<br />
x ≈ 3.05 or x ≈ −0.52<br />
29. First write − 5.4 x( x + 9.8) + 4.1 = 3.2 − 6.9x<br />
in the<br />
2<br />
form ax + bx + c = 0 :<br />
2<br />
−5.4x − 52.92x + 4.1 = 3.2 − 6.9x<br />
2<br />
5.4x<br />
26.02x<br />
0.9 0<br />
− − + =<br />
a = − 5.4, b = − 46.02, c = 0.9<br />
( 46.02) (<br />
2<br />
46.02) 4( 5.4)( 0.9)<br />
2( −5.4)<br />
− − ± − − −<br />
x =<br />
46.02 ± 2137.2804<br />
x =<br />
−10.8<br />
46.02 + 2137.28 46.02 − 2137.28<br />
x ≈<br />
or x ≈<br />
−10.8 −10.8<br />
x ≈ −8.54 or x ≈ 0.020<br />
31. x 2 = 25<br />
2<br />
x − 25 = 0<br />
( x + 5)( x − 5) = 0<br />
x + 5 = 0 or x − 5 = 0<br />
x = − 5 or x = 5<br />
195
Homework 7.3<br />
SSM: Intermediate Algebra<br />
33. x( x − 8) = − 16<br />
2<br />
x − 8x<br />
= −16<br />
2<br />
x − 8x<br />
+ 16 = 0<br />
( x − 4)( x − 4) = 0<br />
x − 4 = 0<br />
x = 4<br />
35. 4x 2 = 80<br />
37.<br />
x<br />
2<br />
= 20<br />
x = ±<br />
20<br />
x = ± 2 5<br />
2<br />
(4x + 3) + 2 = 22<br />
( x ) 2<br />
4 + 3 = 20<br />
4x<br />
+ 3 = ± 20<br />
4x<br />
+ 3 = ± 2 5<br />
4x<br />
= − 3 ± 2 5<br />
x =<br />
− 3 ± 2 5<br />
4<br />
39. 2 x(3x − 4) = x − 5<br />
2<br />
6x 8x x 5<br />
2<br />
6x<br />
9x<br />
5 0<br />
a = 6, b = − 9, c = −5<br />
− − ± − − −<br />
x =<br />
2(6)<br />
x =<br />
− = +<br />
− − =<br />
2<br />
( 9) ( 9) 4(6)( 5)<br />
9 ± −39<br />
12<br />
Since the square root of a negative number is not a<br />
real number, there are no real solutions.<br />
41. 9x<br />
2 − 5x<br />
= 0<br />
x(9x<br />
− 5) = 0<br />
x = 0 or 9x<br />
− 5 = 0<br />
x = 0 or 9x<br />
= 5<br />
5<br />
x = 0 or x =<br />
9<br />
43. 5x<br />
2 = 3x<br />
+ 8<br />
2<br />
5x<br />
3x<br />
8 0<br />
− − =<br />
(5x<br />
− 8)( x + 1) = 0<br />
5x<br />
− 8 = 0 or x + 1 = 0<br />
5x<br />
= 8 or x = −1<br />
8<br />
x = or x = − 1<br />
5<br />
45. 3x<br />
2 = 27x<br />
47.<br />
− =<br />
2<br />
3x<br />
27x<br />
0<br />
3 x( x − 9) = 0<br />
x( x − 9) = 0<br />
x = 0 or x − 9 = 0<br />
x = 0 or x = 9<br />
2<br />
2(2x<br />
1) 3<br />
+ = x :<br />
+ =<br />
2<br />
4x<br />
2 3<br />
x<br />
− + =<br />
2<br />
4x<br />
3x<br />
2 0<br />
a = 4, b = − 3, c = 2<br />
− − ± − −<br />
x =<br />
2(4)<br />
x =<br />
2<br />
( 3) ( 3) 4(4)(2)<br />
3 ± −23<br />
8<br />
Since the square root of a negative number is not a<br />
real number, there are no real solutions.<br />
49. x<br />
2 + 12x<br />
+ 36 = 0<br />
( x + 6)( x + 6) = 0<br />
51.<br />
x + 6 = 0<br />
x = −6<br />
− − =<br />
2<br />
24x<br />
18x<br />
60 0<br />
− − =<br />
2<br />
6(4x<br />
3x<br />
10) 0<br />
− − =<br />
2<br />
4x<br />
3x<br />
10 0<br />
(4x<br />
+ 5)( x − 2) = 0<br />
4x<br />
+ 5 = 0 or x − 2 = 0<br />
4x<br />
= − 5 or x = 2<br />
5<br />
x = − or x = 2<br />
4<br />
196
SSM: Intermediate Algebra Homework 7.3<br />
53.<br />
2<br />
( x − 4) − 3 = 7<br />
− =<br />
2<br />
( x 4) 10<br />
x − 4 = ± 10<br />
x = 4 ± 10<br />
55. x( x + 1) = 7 − x<br />
+ = −<br />
2<br />
x x 7 x<br />
2<br />
x + 2x<br />
= 7<br />
2<br />
x + 2x<br />
+ 1 = 7 + 1<br />
+ =<br />
2<br />
( x 1) 8<br />
x + 1 = ± 8<br />
x + 1 = ± 2 2<br />
x = − 1±<br />
2 2<br />
57. 2 x(3x − 1) = 3( x − 2)<br />
61.<br />
63.<br />
3 169<br />
x − = ±<br />
2 4<br />
3 13<br />
x − = ±<br />
2 2<br />
3 13<br />
x = ±<br />
2 2<br />
3 13 3 13<br />
x = + or x = −<br />
2 2 2 2<br />
16 −10<br />
x = =8 or x = = −5<br />
2 2<br />
=<br />
2<br />
5x<br />
35<br />
x<br />
− =<br />
2<br />
5x<br />
35x<br />
0<br />
5 x( x − 7) = 0<br />
5x<br />
= 0 or x − 7 = 0<br />
x = 0 or x = 7<br />
= +<br />
2<br />
50x<br />
5x<br />
105<br />
− = −<br />
2<br />
6x 2x 3x<br />
6<br />
− + =<br />
2<br />
6x<br />
5x<br />
6 0<br />
a = 6, b = − 5, c = 6<br />
2<br />
( 5) ( 5) 4(6)(6)<br />
− − ± − −<br />
x =<br />
2(6)<br />
x =<br />
5 ± −119<br />
12<br />
Since the square root of a negative number is not a<br />
real number, there are no real solutions.<br />
59. 2 x(3x − 1) = 3( x − 2)<br />
− =<br />
2<br />
2x<br />
6x<br />
80<br />
− =<br />
2<br />
2( x 3 x) 80<br />
2<br />
x − 3x<br />
= 40<br />
2 9 9<br />
x − 3x<br />
+ = 40 +<br />
4 4<br />
2<br />
⎛ 3 ⎞ 160 9<br />
⎜ x − ⎟ = +<br />
⎝ 2 ⎠ 4 4<br />
⎛ 3 ⎞ 169<br />
⎜ x − ⎟ =<br />
⎝ 2 ⎠ 4<br />
2<br />
65.<br />
− + =<br />
2<br />
5x<br />
50x<br />
105 0<br />
− + =<br />
2<br />
5( x 25x<br />
21) 0<br />
2<br />
x − 25x<br />
+ 21 = 0<br />
( x − 3)( x − 7) = 0<br />
x − 3 = 0 or x − 7 = 0<br />
x = 3 or x = 7<br />
2<br />
(3x − 1) = 7<br />
3x<br />
− 1 = ± 7<br />
3x<br />
= 1±<br />
7<br />
x =<br />
1±<br />
7<br />
3<br />
67. ( x − 2)( x − 5) = 1<br />
− 5 − 2 + 10 = 1<br />
2<br />
x x x<br />
2<br />
x − 7x<br />
= −9<br />
2 49 49<br />
x − 7x<br />
+ = − 9 +<br />
4 4<br />
2<br />
⎛ 7 ⎞ 36 49<br />
⎜ x − ⎟ = − +<br />
⎝ 2 ⎠ 4 4<br />
⎛ 7 ⎞ 13<br />
⎜ x − ⎟ =<br />
⎝ 2 ⎠ 4<br />
2<br />
197
Homework 7.3<br />
SSM: Intermediate Algebra<br />
69.<br />
71.<br />
7 13<br />
x − = ±<br />
2 4<br />
7 13<br />
x − = ±<br />
2 2<br />
7 13<br />
x = ±<br />
2 2<br />
x =<br />
7 ± 13<br />
2<br />
2 2<br />
( x 1) ( x 2) 0<br />
− + + =<br />
− 2 + 1+ + 4 + 4 = 0<br />
2 2<br />
x x x x<br />
+ + =<br />
2<br />
2x<br />
2x<br />
5 0<br />
2 5<br />
x + x + = 0<br />
2<br />
2 5<br />
x + x = −<br />
2<br />
2 1 5 1<br />
x + x + = − +<br />
4 2 4<br />
2<br />
⎛ 1 ⎞ 10 1<br />
⎜ x + ⎟ = − +<br />
⎝ 2 ⎠ 4 4<br />
⎛ 1 ⎞ 9<br />
⎜ x + ⎟ = −<br />
⎝ 2 ⎠ 4<br />
1 9<br />
x + = ± −<br />
2 4<br />
2<br />
Since the square root of a negative number is not a<br />
real number, there are no real solutions.<br />
2 2<br />
( x 2)( x 5) x 5x 2x 10 x 3x<br />
10<br />
+ − = − + − = − −<br />
73. ( x + 2)( x − 5) = 3<br />
2<br />
x − 3x<br />
− 10 = 3<br />
2<br />
x − 3x<br />
− 13 = 0<br />
a = 1, b = − 3, c = −13<br />
− − ± − − −<br />
x =<br />
2(1)<br />
2<br />
( 3) ( 3) 4(1)( 13)<br />
3 ± 61<br />
x =<br />
2<br />
75.<br />
77.<br />
2<br />
4( x 2) 3 1<br />
− − + = −<br />
−4( x − 2)( x − 2) + 3 = −1<br />
− − − + + = −<br />
2<br />
4( x 2x 2x<br />
4) 3 1<br />
− − + + = −<br />
2<br />
4( x 4x<br />
4) 3 1<br />
− + − + = −<br />
2<br />
4x<br />
16x<br />
16 3 1<br />
− + − =<br />
2<br />
4x<br />
16x<br />
12 0<br />
− − + =<br />
2<br />
4( x 4x<br />
3) 0<br />
2<br />
x − 4x<br />
+ 3 = 0<br />
( x − 3)( x − 1) = 0<br />
x − 3 = 0 or x − 1 = 0<br />
x = 3 or x = 1<br />
− − + = − − − +<br />
2<br />
4( x 2) 3 4( x 2)( x 2) 3<br />
= − − − + +<br />
2<br />
4( x 2x 2x<br />
4) 3<br />
= − − + +<br />
2<br />
4( x 4x<br />
4) 3<br />
= − + − +<br />
2<br />
4x<br />
16x<br />
16 3<br />
= − + −<br />
2<br />
4x<br />
16x<br />
13<br />
79. a. Substitute 3 for f(x):<br />
2<br />
3 x 4x<br />
8<br />
= − +<br />
2<br />
x − 4x<br />
+ 5 = 0<br />
a = 1, b = − 4, c = 5<br />
2 2<br />
b − 4 ac = ( −4) − 4(1)(5) = 16 − 20 = − 4 < 0<br />
So, there are no real number solutions, which<br />
means there are no points on f at y = 3.<br />
b. Substitute 4 for f(x):<br />
= − +<br />
2<br />
4 x 4x<br />
8<br />
2<br />
x − 4x<br />
+ 4 = 0<br />
a = 1, b = − 4, c = 4<br />
2 2<br />
b − 4 ac = ( −4) − 4(1)(4) = 16 − 16 = 0<br />
So, there is one solution <strong>to</strong> the equation, which<br />
means there is one point on f at y = 4.<br />
198
SSM: Intermediate Algebra Homework 7.3<br />
c. Substitute 5 for f(x):<br />
83. a.<br />
= − +<br />
2<br />
5 x 4x<br />
8<br />
2<br />
x − 4x<br />
+ 3 = 0<br />
a = 1, b = − 4, c = 3<br />
2 2<br />
b − 4 ac = ( −4) − 4(1)(3) = 16 − 12 = 4 > 0<br />
d.<br />
So, there are two real number solutions, which<br />
means there are two points on f at y = 5.<br />
Yes, f does model the data well.<br />
b. In 2007, t = 17. Solve for f(17)<br />
f = − +<br />
2<br />
(17) 30.32(17) 122.15(17) 109.75<br />
= 6795.68<br />
≈ 6796 schools<br />
c. 200 schools per state means 10,000 charter<br />
schools. Solve for t when f(t) = 10,000.<br />
81. Solve for x when f(x) = 2:<br />
= − +<br />
2<br />
2 x 6x<br />
7<br />
2<br />
x − 6x<br />
+ 5 = 0<br />
( x − 5)( x − 1) = 0<br />
x − 5 = 0 or x − 1 = 0<br />
x = 5 or x = 1<br />
Therefore, two points at height 2 are ( 1,2 ) and<br />
( 5,2 ) . Since these points are symmetric and the<br />
average of the x-coordinates at these points is<br />
1 + 5 = 3 the x-coordinate of the vertex is 3.<br />
2<br />
2<br />
Substitute 3 for x in f ( x) = x − 6x<br />
+ 7 <strong>to</strong> find the<br />
y-coordinate of the vertex:<br />
2<br />
f (3) = 3 − 6(3) + 7 = − 2 So the vertex is ( 3, − 2)<br />
.<br />
85. a.<br />
2<br />
10000 30.32t<br />
122.15t<br />
109.75<br />
2<br />
0 30.32t<br />
122.15t<br />
9890.25<br />
t =<br />
= − +<br />
= − −<br />
2<br />
122.15 (122.15) 4(30.32)( 9890.25)<br />
t ≈ −16.16 or t ≈ 20.19<br />
t ≈ −16 or t ≈ 20<br />
± − −<br />
2(30.32)<br />
The solution t = -16 is model breakdown.<br />
There will be 200 charter schools per state in<br />
2010.<br />
b.<br />
Yes, f does model the data well.<br />
2<br />
f (13) = 23.43(13) − 259.14(13) + 815.77<br />
≈ 1406.65<br />
≈ 1407<br />
This means that in 2003, there will be1407<br />
convictions of police officers.<br />
199
Homework 7.3<br />
SSM: Intermediate Algebra<br />
c.<br />
= − +<br />
2<br />
2000 23.43t<br />
259.14t<br />
815.77<br />
2<br />
0 23.43t<br />
259.14t<br />
815.77<br />
2<br />
0 23.43t<br />
259.14t<br />
1184.23<br />
t =<br />
= − +<br />
= − −<br />
2<br />
259.14 (259.14) 4(23.43)( 1184.23)<br />
2(23.43)<br />
t ≈ −3.48 or t ≈ 14.54<br />
t ≈ −3 or t ≈ 15<br />
± − −<br />
There were 2000 convictions of police officers<br />
in 1987 and 2005.<br />
Since the square root of a negative number is not a<br />
real number, there are no real number solutions,<br />
and therefore, no x-intercepts.<br />
93. a. mx + b = 0<br />
mx + b − b = 0 − b<br />
mx = −b<br />
mx b<br />
= −<br />
m m<br />
b<br />
x = −<br />
m<br />
87. No, the student did not solve the equation correctly<br />
because they did not change the form in<strong>to</strong><br />
2<br />
ax + bx + c = 0 first. Here is the correct way:<br />
2<br />
2x<br />
5x<br />
1<br />
2<br />
2x<br />
5x<br />
1 0<br />
So, a = 2, b = 5, c = −1<br />
− ± − −<br />
x =<br />
2(2)<br />
x =<br />
+ =<br />
+ − =<br />
2<br />
5 (5) 4(2)( 1)<br />
− 5 ± 33<br />
4<br />
89. Solve for x when f(x) = 0:<br />
2<br />
0 x 4x<br />
6<br />
2<br />
( 4) ( 4) 4(1)( 6)<br />
− − ± − − −<br />
x =<br />
2(1)<br />
x =<br />
x =<br />
= − −<br />
4 ± 40<br />
2<br />
4 ± 2 10<br />
2<br />
x = 2 ± 10<br />
x ≈ 5.16 or x ≈ −1.16<br />
The x-intercepts are (5.16, 0) and (-1.16, 0)<br />
91. Solve for x when h(x) = 0:<br />
2<br />
0 3x<br />
2x<br />
5<br />
− − ± − −<br />
x =<br />
2(3)<br />
x =<br />
= − +<br />
2<br />
( 2) ( 2) 4(3)(5)<br />
2 ± −56<br />
6<br />
b. 7x + 21 = 0<br />
So, m = 7 and b = -21. <strong>Using</strong> the formula from<br />
part a:<br />
21<br />
x = − = − 3<br />
7<br />
Solving for x in the usual way:<br />
7x<br />
+ 21 = 0<br />
7x<br />
+ 21− 21 = 0 − 21<br />
7x<br />
= −21<br />
7x<br />
21<br />
= −<br />
7 7<br />
x = −3<br />
95. a = k, b = -12, c = 4<br />
2<br />
b − 4ac<br />
= 0<br />
− − =<br />
2<br />
( 12) 4( k)(4) 0<br />
144 − 16k<br />
= 0<br />
97. a = 2, b = k, c = 8<br />
k<br />
2<br />
2<br />
b − 4ac<br />
= 0<br />
− 4(2)(8) = 0<br />
k<br />
2<br />
− 16k<br />
= −144<br />
− 64 = 0<br />
k<br />
2<br />
k = 9<br />
= 64<br />
k = ± 8<br />
200
SSM: Intermediate Algebra Homework 7.4<br />
99. a = 9, b = -6, c = k<br />
2<br />
b − 4ac<br />
= 0<br />
− − =<br />
2<br />
( 6) 4(9)( k) 0<br />
36 − 36k<br />
= 0<br />
− 36k<br />
= −36<br />
k = 1<br />
101. Fac<strong>to</strong>ring:<br />
2<br />
x − x − 20 = 0<br />
( x − 5)( x + 4) = 0<br />
x − 5 = 0 or x + 4 = 0<br />
x = 5 or x = −4<br />
Completing the square:<br />
2<br />
x − x = 20<br />
2 1 1<br />
x − x + = 20 +<br />
4 4<br />
⎛ 1 ⎞ 81<br />
⎜ x − ⎟ =<br />
⎝ 2 ⎠ 4<br />
2<br />
1 81<br />
x − = ±<br />
2 4<br />
1 9<br />
x − = ±<br />
2 2<br />
1 9 1 9<br />
x = + or x = −<br />
2 2 2 2<br />
10 8<br />
x = or x = −<br />
2 2<br />
x = 5 or x = −4<br />
<strong>Quadratic</strong> formula:<br />
2<br />
x − x − 20 = 0<br />
− − ± − − −<br />
x =<br />
2(1)<br />
2<br />
( 1) ( 1) 4(1)( 20)<br />
1±<br />
81<br />
x =<br />
2<br />
1±<br />
9<br />
x =<br />
2<br />
1+ 9 1−<br />
9<br />
x = or x =<br />
2 2<br />
x = 5 or x = −4<br />
103. Answers may vary. Example:<br />
2<br />
First, put the equation in the form ax + bx + c = 0 .<br />
Once you know the value of a, b, and c, substitute<br />
these values in<strong>to</strong> the quadratic formula.<br />
− ±<br />
x =<br />
Homework 7.4<br />
−<br />
2a<br />
2<br />
b b 4ac<br />
1. Substitute the given points in<strong>to</strong><br />
2<br />
( 1,6 ): 6 = a( 1) + b( 1)<br />
+ c<br />
( 2,11 ):11 2<br />
( 2) ( 2)<br />
( 3,8 ): 8 =<br />
2<br />
( 3) + ( 3)<br />
+<br />
= a + b + c<br />
a b c<br />
Simplify these equations:<br />
a+ b+ c = 6 1<br />
( )<br />
( )<br />
( )<br />
4a + 2b + c = 11 2<br />
9a + 3b + c = 18 3<br />
2<br />
y = ax + bx + c .<br />
Eliminate c by multiplying both sides of equation<br />
(1) by -1:<br />
−a − b − c = − 6 4<br />
( )<br />
Adding the left sides and right sides of equations<br />
(2) and (4) gives:<br />
3a<br />
+ b = 5 5<br />
( )<br />
Adding the left sides and right sides of equations<br />
(3) and (4) gives:<br />
8a<br />
+ 2b<br />
= 12 6<br />
Simplify:<br />
4a<br />
+ b = 6 7<br />
( )<br />
( )<br />
Eliminate b by multiplying equation (5) by -1 and<br />
add each side <strong>to</strong> the corresponding side of equation<br />
(7):<br />
a = 1<br />
Next, substitute 1 for a in equation (5):<br />
3 1 + b = 5<br />
( )<br />
b = 2<br />
Then, substitute 1 for a and 2 for b in equation (1):<br />
a+ b+ c = 6<br />
1+ 2 + c = 6<br />
c = 3<br />
Therefore, a = 1, b = 2, and c = 3. So, the equation<br />
2<br />
is y = x + 2x<br />
+ 3 .<br />
201
Homework 7.4<br />
SSM: Intermediate Algebra<br />
3. Substitute the given points in<strong>to</strong><br />
2<br />
( 1,5 ): 5 = a( 1) + b( 1)<br />
+ c<br />
( 2,11 ):11 2<br />
( 2) ( 2)<br />
( 3,19 ):19 2<br />
( 3) ( 3)<br />
= a + b + c<br />
= a + b + c<br />
Simplify these equations:<br />
a+ b+ c = 5 1<br />
( )<br />
( )<br />
( )<br />
4a + 2b + c = 11 2<br />
9a + 3b + c = 19 3<br />
2<br />
y = ax + bx + c .<br />
Eliminate c by multiplying both sides of equation<br />
(1) by -1:<br />
−a − b − c = − 5 4<br />
( )<br />
Adding the left sides and right sides of equations<br />
(2) and (4) gives:<br />
3a<br />
+ b = 6 5<br />
( )<br />
Adding the left sides and right sides of equations<br />
(3) and (4) gives:<br />
8a<br />
+ 2b<br />
= 14 6<br />
Simplify:<br />
4a<br />
+ b = 7 7<br />
( )<br />
( )<br />
Eliminate b by multiplying equation (5) by -1 and<br />
add each side <strong>to</strong> the corresponding side of equation<br />
(7):<br />
a = 1<br />
Next, substitute 1 for a in equation (5):<br />
3 1 + b = 6<br />
( )<br />
b = 3<br />
Then, substitute 1 for a and 3 for b in equation (1):<br />
a+ b+ c = 5<br />
1+ 3 + c = 5<br />
c = 1<br />
Therefore, a = 1, b = 3, and c = 1. So, the equation<br />
2<br />
is y = x + 3x<br />
+ 1.<br />
5. Substitute the given points in<strong>to</strong><br />
2<br />
( 1,9 ): 9 = a( 1) + b( 1)<br />
+ c<br />
2<br />
( 2,7 ): 7 = a( 2) + b( 2)<br />
+ c<br />
2<br />
( 4, 15 ): 15 ( 4) ( 4)<br />
− − = a + b + c<br />
Simplify these equations:<br />
a+ b+ c = 9 1<br />
( )<br />
( )<br />
( )<br />
4a + 2b + c = 7 2<br />
16a + 4b + c = −15 3<br />
2<br />
y = ax + bx + c .<br />
Eliminate c by multiplying both sides of equation<br />
(1) by -1:<br />
( )<br />
−a − b − c = − 9 4<br />
Adding the left sides and right sides of equations<br />
(2) and (4) gives:<br />
3a<br />
+ b = − 2 5<br />
( )<br />
Adding the left sides and right sides of equations<br />
(3) and (4) gives:<br />
15a<br />
+ 3b<br />
= − 24 6<br />
Simplify:<br />
5a<br />
+ b = − 8 7<br />
( )<br />
( )<br />
Eliminate b by multiplying equation (5) by -1 and<br />
add each side <strong>to</strong> the corresponding side of equation<br />
(7):<br />
2a<br />
= −6<br />
a = −3<br />
Next, substitute -3 for a in equation (5):<br />
3 − 3 + b = −2<br />
( )<br />
b = 7<br />
Then, substitute -3 for a and 7 for b in equation (1):<br />
a+ b+ c = 9<br />
− 3 + 7 + c = 9<br />
c = 5<br />
Therefore, a = -3, b = 7, and c = 5. So, the equation<br />
2<br />
is y = − 3x + 7x<br />
+ 5 .<br />
7. Substitute the given points in<strong>to</strong><br />
2<br />
( 2, 2 ): 2 = a( 2) + b( 2)<br />
+ c<br />
( 3,11 ):11 =<br />
2<br />
( 3) + ( 3)<br />
+<br />
( 4, 24 ): 24<br />
2<br />
( 4) ( 4)<br />
a b c<br />
= a + b + c<br />
Simplify these equations:<br />
a+ b+ c = 2 1<br />
( )<br />
( )<br />
( )<br />
9a + 3b + c = 11 2<br />
16a + 4b + c = 24 3<br />
2<br />
y = ax + bx + c .<br />
Eliminate c by multiplying both sides of equation<br />
(1) by -1:<br />
−4a − 2b − c = − 2 4<br />
( )<br />
Adding the left sides and right sides of equations<br />
(2) and (4) gives:<br />
5a<br />
+ b = 9 5<br />
( )<br />
Adding the left sides and right sides of equations<br />
(3) and (4) gives:<br />
12a<br />
+ 2b<br />
= 22 6<br />
Simplify:<br />
6a<br />
+ b = 11 7<br />
( )<br />
( )<br />
202
SSM: Intermediate Algebra Homework 7.4<br />
Eliminate b by multiplying equation (5) by -1 and<br />
add each side <strong>to</strong> the corresponding side of equation<br />
(7):<br />
a = 2<br />
Next, substitute 2 for a in equation (5):<br />
5 2 + b = 9<br />
( )<br />
b = −1<br />
Then, substitute 2 for a and -1 for b in equation (1):<br />
4a + 2b + c = 2<br />
4(2) + 2( − 1) + c = 2<br />
8 − 2 + c = 2<br />
c = −4<br />
Therefore, a = 2, b = -1, and c = -4. So, the<br />
2<br />
equation is y = 2x − x − 4 .<br />
9. Substitute the given points in<strong>to</strong><br />
2<br />
( 1, −3 ): − 3 = ( 1) + ( 1)<br />
+<br />
2<br />
( 3,9 ): 9 = a( 3) + b( 3)<br />
+ c<br />
2<br />
( 5,29 ): 29 ( 5) ( 5)<br />
a b c<br />
= a + b + c<br />
Simplify these equations:<br />
a+ b+ c = −3 1<br />
( )<br />
( )<br />
( )<br />
9a + 3b + c = 9 2<br />
25a + 5b + c = 29 3<br />
2<br />
y = ax + bx + c .<br />
Eliminate c by multiplying both sides of equation<br />
(1) by -1:<br />
−a − b − c = 3 4<br />
( )<br />
Adding the left sides and right sides of equations<br />
(2) and (4) gives:<br />
8a<br />
+ 2b<br />
= 12 5<br />
Simplify:<br />
4a<br />
+ b = 6 6<br />
( )<br />
( )<br />
Adding the left sides and right sides of equations<br />
(3) and (4) gives:<br />
24a<br />
+ 4b<br />
= 32 7<br />
Simplify:<br />
6a<br />
+ b = 8 8<br />
( )<br />
( )<br />
Eliminate b by multiplying equation (6) by -1 and<br />
add each side <strong>to</strong> the corresponding side of equation<br />
(8):<br />
2a<br />
= 2<br />
a = 1<br />
Next, substitute 1 for a in equation (6):<br />
4(1) + b = 6<br />
b = 2<br />
Then, substitute 1 for a and 2 for b in equation (1):<br />
a+ b+ c = −3<br />
1+ 2 + c = −3<br />
c = −6<br />
Therefore, a = 1, b = 2, and c = -6. So, the equation<br />
2<br />
is y = x + 2x<br />
− 6 .<br />
11. Substitute the given points in<strong>to</strong><br />
2<br />
( 3,7 ): 7 = a( 3) + b( 3)<br />
+ c<br />
2<br />
( 4,0 ): 0 = a( 4) + b( 4)<br />
+ c<br />
2<br />
( 5, 11 ): 11 ( 5) ( 5)<br />
− − = a + b + c<br />
Simplify these equations:<br />
9a + 3b + c = 7 1<br />
( )<br />
( )<br />
( )<br />
16a + 4b + c = 0 2<br />
25a + 5b + c = −11 3<br />
2<br />
y = ax + bx + c .<br />
Eliminate c by multiplying both sides of equation<br />
(1) by -1:<br />
−9a − 3b − c = − 7 4<br />
( )<br />
Adding the left sides and right sides of equations<br />
(2) and (4) gives:<br />
7a<br />
+ b = − 7 5<br />
( )<br />
Adding the left sides and right sides of equations<br />
(3) and (4) gives:<br />
16a<br />
+ 2b<br />
= − 18 6<br />
Simplify:<br />
8a<br />
+ b = − 9 7<br />
( )<br />
( )<br />
Eliminate b by multiplying equation (5) by -1 and<br />
add each side <strong>to</strong> the corresponding side of equation<br />
(7):<br />
a = − 2<br />
Next, substitute -2 for a in equation (5):<br />
7 − 2 + b = −7<br />
( )<br />
b = 7<br />
Then, substitute -2 for a and 7 for b in equation (1):<br />
9a + 3b + c = 7<br />
9( − 2) + 3(7) + c = 7<br />
− 18 + 21+ c = 7<br />
c = 4<br />
Therefore, a = -2, b = 7, and c = 4. So, the equation<br />
2<br />
is y = − 2x + 7x<br />
+ 4 .<br />
203
Homework 7.4<br />
SSM: Intermediate Algebra<br />
13. Substitute the given points in<strong>to</strong><br />
( 3,2 ): 2 =<br />
2<br />
( 3) + ( 3)<br />
+<br />
( 4,16 ):16 2<br />
( 4) ( 4)<br />
( 5,36 ): 36<br />
2<br />
( 5) ( 5)<br />
a b c<br />
= a + b + c<br />
= a + b + c<br />
Simplify these equations:<br />
9a + 3b + c = 2 1<br />
( )<br />
( )<br />
( )<br />
16a + 4b + c = 16 2<br />
25a + 5b + c = 36 3<br />
2<br />
y = ax + bx + c .<br />
Eliminate c by multiplying both sides of equation<br />
(1) by -1:<br />
−9a − 3b − c = − 2 4<br />
( )<br />
Adding the left sides and right sides of equations<br />
(2) and (4) gives:<br />
7a<br />
+ b = 14 5<br />
( )<br />
Adding the left sides and right sides of equations<br />
(3) and (4) gives:<br />
16a<br />
+ 2b<br />
= 34 6<br />
Simplify:<br />
8a<br />
+ b = 17 7<br />
( )<br />
( )<br />
Eliminate b by multiplying equation (5) by -1 and<br />
add each side <strong>to</strong> the corresponding side of equation<br />
(7):<br />
a = 3<br />
Next, substitute 3 for a in equation (5):<br />
7 3 + b = 14<br />
( )<br />
b = −7<br />
Then, substitute 3 for a and -7 for b in equation (1):<br />
a+ b+ c = 2<br />
3 − 7 + c = 2<br />
c = −4<br />
Therefore, a = 3, b = -7, and c = -4. So, the<br />
2<br />
equation is y = 3x − 7x<br />
− 4 .<br />
15. Substitute the given points in<strong>to</strong><br />
2<br />
( 2, 5 ): 5 ( 2) ( 2)<br />
2<br />
( 4,3 ): 3 = a( 4) + b( 4)<br />
+ c<br />
2<br />
( 5,13 ):13 = ( 5) + ( 5)<br />
+<br />
− − = a + b + c<br />
a b c<br />
Simplify these equations:<br />
4a + 2b + c = −5 1<br />
( )<br />
( )<br />
( )<br />
16a + 4b + c = 3 2<br />
25a + 5b + c = 13 3<br />
2<br />
y = ax + bx + c .<br />
Eliminate c by multiplying both sides of equation<br />
(1) by -1:<br />
( )<br />
−4a − 2b − c = 5 4<br />
Adding the left sides and right sides of equations<br />
(2) and (4) gives:<br />
12a<br />
+ 2b<br />
= 8 5<br />
Simplify:<br />
6a<br />
+ b = 4 6<br />
( )<br />
( )<br />
Adding the left sides and right sides of equations<br />
(3) and (4) gives:<br />
21a<br />
+ 3b<br />
= 18 7<br />
Simplify:<br />
7a<br />
+ b = 6 8<br />
( )<br />
( )<br />
Eliminate b by multiplying equation (6) by -1 and<br />
add each side <strong>to</strong> the corresponding side of equation<br />
(8):<br />
a = 2<br />
Next, substitute 2 for a in equation (6):<br />
6 2 + b = 4<br />
( )<br />
b = −8<br />
Then, substitute 2 for a and -8 for b in equation (1):<br />
4a + 2b + c = −5<br />
4(2) + 2( − 8) + c = −5<br />
8 − 16 = −5<br />
c = 3<br />
Therefore, a = 2, b = -8, and c = 3. So, the equation<br />
2<br />
is y = 2x − 8x<br />
+ 3 .<br />
17. Substitute the given points in<strong>to</strong><br />
( 1, −1 ): − 1 =<br />
2<br />
( 1) + ( 1)<br />
+<br />
( 2, 1 ): 1<br />
2<br />
( 2) ( 2)<br />
2<br />
( 3,3 ): 3 = a( 3) + b( 3)<br />
+ c<br />
a b c<br />
− − = a + b + c<br />
Simplify these equations:<br />
a+ b+ c = −1 1<br />
( )<br />
( )<br />
( )<br />
4a + 2b + c = −1 2<br />
9a + 3b + c = 3 3<br />
2<br />
y = ax + bx + c .<br />
Eliminate c by multiplying both sides of equation<br />
(1) by -1:<br />
−a − b − c = 1 4<br />
( )<br />
Adding the left sides and right sides of equations<br />
(2) and (4) gives:<br />
3a<br />
+ b = 0 5<br />
( )<br />
Adding the left sides and right sides of equations<br />
(3) and (4) gives:<br />
8a<br />
+ 2b<br />
= 4 6<br />
Simplify:<br />
( )<br />
204
SSM: Intermediate Algebra Homework 7.4<br />
( )<br />
4a<br />
+ b = 2 7<br />
Eliminate b by multiplying equation (5) by -1 and<br />
add each side <strong>to</strong> the corresponding side of equation<br />
(7):<br />
a = 2<br />
Next, substitute 2 for a in equation (5):<br />
3 2 + b = 0<br />
( )<br />
b = −6<br />
Then, substitute 2 for a and -6 for b in equation (1):<br />
a+ b+ c = −1<br />
2 − 6 + c = −1<br />
c = 3<br />
Therefore, a = 2, b = -6, and c = 3. So, the equation<br />
2<br />
is y = 2x − 6x<br />
+ 3 .<br />
19. Substitute the given points in<strong>to</strong><br />
( 0, 4 ): 4<br />
2<br />
( 0) ( 0)<br />
( 2,8 ): 8<br />
2<br />
( 2) ( 2)<br />
( 3,1 ):1 =<br />
2<br />
( 3) + ( 3)<br />
+<br />
= a + b + c<br />
= a + b + c<br />
a b c<br />
Simplify these equations:<br />
c = 4 1<br />
( )<br />
( )<br />
( )<br />
4a + 2b + c = 8 2<br />
9a + 3b + c = 1 3<br />
2<br />
y = ax + bx + c .<br />
Since c = 4, substitute 4 for c in equations (2) and<br />
(3):<br />
4a<br />
+ 2b<br />
+ 4 = 8<br />
9a<br />
+ 3b<br />
+ 4 = 1<br />
Simplifying these equations gives:<br />
4a<br />
+ 2b<br />
= 4 4<br />
( )<br />
( )<br />
9a<br />
+ 3b<br />
= −3 5<br />
To eliminate b, multiply both sides of equation (4)<br />
by -3 and both sides of equation (5) by 2:<br />
−12a<br />
− 6b<br />
= −12 6<br />
( )<br />
( )<br />
18a<br />
+ 6b<br />
= −6 7<br />
Adding the left and the right sides of equations (6)<br />
and (7) gives:<br />
6a<br />
= −18<br />
a = −3<br />
Next, substitute -3 for a in equation (4):<br />
4( − 3) + 2b<br />
= 4<br />
2b<br />
= 16<br />
b = 8<br />
Therefore, a = -3, b = 8, and c = 4. So, the equation<br />
2<br />
is y = − 3x + 8x<br />
+ 4 .<br />
21. Substitute the given points in<strong>to</strong><br />
2<br />
( 0, 1 ): 1 ( 0) ( 0)<br />
2<br />
( 1,3 ): 3 = a( 1) + b( 1)<br />
+ c<br />
2<br />
( 2,13 ):13 ( 2) ( 2)<br />
− − = a + b + c<br />
= a + b + c<br />
Simplify these equations:<br />
c = −1 1<br />
( )<br />
( )<br />
( )<br />
a+ b+ c = 3 2<br />
4a + 2b + c = 13 3<br />
2<br />
y = ax + bx + c .<br />
Since c = -1, substitute -1 for c in equations (2) and<br />
(3):<br />
a + b + ( − 1) = 3<br />
4a<br />
+ 2 b + ( − 1) = 13<br />
Simplifying these equations gives:<br />
a + b = 4 4<br />
( )<br />
( )<br />
4a<br />
+ 2b<br />
= 14 5<br />
To eliminate b, multiply both sides of equation (4)<br />
by -2 and add each side <strong>to</strong> the corresponding side<br />
in equation (5):<br />
2a<br />
= 6<br />
a = 3<br />
Next, substitute 3 for a in equation (4):<br />
3 + b = 4<br />
b = 1<br />
Therefore, a = 3, b = 1, and c = -1. So, the<br />
2<br />
equation is y = 3x + x − 1.<br />
23. Substitute the given points in<strong>to</strong><br />
( 0,17 ):17 2<br />
( 0) ( 0)<br />
( 2,11 ):11 2<br />
( 2) ( 2)<br />
( 3,2 ): 2 =<br />
2<br />
( 3) + ( 3)<br />
+<br />
= a + b + c<br />
= a + b + c<br />
a b c<br />
Simplify these equations:<br />
c = 17 1<br />
( )<br />
( )<br />
( )<br />
4a + 2b + c = 11 2<br />
9a + 3b + c = 2 3<br />
2<br />
y = ax + bx + c .<br />
Since c = 17, substitute 17 for c in equations (2)<br />
and (3):<br />
4a<br />
+ 2b<br />
+ 17 = 11<br />
9a<br />
+ 3b<br />
+ 17 = 2<br />
Simplifying these equations gives:<br />
4a<br />
+ 2b<br />
= −6 4<br />
( )<br />
( )<br />
9a<br />
+ 3b<br />
= −15 5<br />
To eliminate b, multiply both sides of equation (4)<br />
by -3 and both sides of equation (5) by 2:<br />
205
Homework 7.4<br />
SSM: Intermediate Algebra<br />
( )<br />
( )<br />
−12a<br />
− 6b<br />
= 18 6<br />
18a<br />
+ 6b<br />
= −30 7<br />
Adding the left and the right sides of equations (6)<br />
and (7) gives:<br />
6a<br />
= −12<br />
a = −2<br />
Next, substitute -2 for a in equation (4):<br />
4( − 2) + 2b<br />
= −6<br />
2b<br />
= 2<br />
b = 1<br />
Therefore, a = -2, b = 1, and c = 17. So, the<br />
2<br />
equation is y = − 2x + x + 17 .<br />
25. Substitute the given points in<strong>to</strong><br />
( 1,1 ):1 =<br />
2<br />
( 1) + ( 1)<br />
+<br />
( 2, 4 ): 4<br />
2<br />
( 2) ( 2)<br />
( 3,9 ): 9<br />
2<br />
( 3) ( 3)<br />
a b c<br />
= a + b + c<br />
= a + b + c<br />
Simplify these equations:<br />
a+ b+ c = 1 1<br />
( )<br />
( )<br />
( )<br />
4a + 2b + c = 4 2<br />
9a + 3b + c = 9 3<br />
2<br />
y = ax + bx + c .<br />
Eliminate c by multiplying both sides of equation<br />
(1) by -1:<br />
−a − b − c = − 1 4<br />
( )<br />
Adding the left sides and right sides of equations<br />
(2) and (4) gives:<br />
3a<br />
+ b = 3 5<br />
( )<br />
Adding the left sides and right sides of equations<br />
(3) and (4) gives:<br />
8a<br />
+ 2b<br />
= 8 6<br />
( )<br />
Simplify:<br />
4a<br />
+ b = 4 7<br />
( )<br />
Eliminate b by multiplying equation (5) by -1 and<br />
add each side <strong>to</strong> the corresponding side of equation<br />
(7):<br />
a = 1<br />
Next, substitute 1 for a in equation (5):<br />
3 1 + b = 3<br />
( )<br />
b = 0<br />
Then, substitute 1 for a and 0 for b in equation (1):<br />
a+ b+ c = 1<br />
1+ 0 + c = 1<br />
c = 0<br />
Therefore, a = 1, b = 0, and c = 0. So, the equation<br />
2<br />
is y = x .<br />
27. Substitute the given points in<strong>to</strong><br />
2<br />
f ( x)<br />
= ax + bx + c .<br />
( 1,1 ):1 =<br />
2<br />
( 1) + ( 1)<br />
+<br />
( 2, 2 ): 2<br />
2<br />
( 2) ( 2)<br />
( 3,3 ): 3<br />
2<br />
( 3) ( 3)<br />
a b c<br />
= a + b + c<br />
= a + b + c<br />
Simplify these equations:<br />
a+ b+ c = 1 1<br />
( )<br />
( )<br />
( )<br />
4a + 2b + c = 2 2<br />
9a + 3b + c = 3 3<br />
Eliminate c by multiplying both sides of equation<br />
(1) by -1:<br />
−a − b − c = − 1 4<br />
( )<br />
Adding the left sides and right sides of equations<br />
(2) and (4) gives:<br />
3a<br />
+ b = 1 5<br />
( )<br />
Adding the left sides and right sides of equations<br />
(3) and (4) gives:<br />
8a<br />
+ 2b<br />
= 2 6<br />
Simplify:<br />
4a<br />
+ b = 1 7<br />
( )<br />
( )<br />
Eliminate b by multiplying equation (5) by -1 and<br />
add each side <strong>to</strong> the corresponding side of equation<br />
(7):<br />
a = 0<br />
Next, substitute 0 for a in equation (5):<br />
3 0 + b = 1<br />
( )<br />
b = 1<br />
Then, substitute 0 for a and 1 for b in equation (1):<br />
a+ b+ c = 1<br />
0 + 1+ c = 1<br />
c = 0<br />
Therefore, a = 0, b = 1, and c = 0. So, the equation<br />
f x = x , which is a linear function.<br />
is ( )<br />
2<br />
30. Solve for a in y = a( x − h)<br />
+ k by substituting 5<br />
for h and -7 for k since (5, -7) is the vertex, and<br />
substitute 8 for x and 11 for y since (8, 11) lies on<br />
the parabola:<br />
206
SSM: Intermediate Algebra Homework 7.4<br />
= − +<br />
2<br />
y a( x h)<br />
k<br />
= − + −<br />
2<br />
11 a(8 5) ( 7)<br />
11 = 9a<br />
− 7<br />
18 = 9a<br />
a = 2<br />
2<br />
Since a = 2, y = 2( x − 5) − 7 or, expanding the<br />
solution:<br />
y = 2( x − 5)( x − 5) − 7<br />
= − + −<br />
2<br />
2( x 10x<br />
25) 7<br />
= − + −<br />
2<br />
2x<br />
20x<br />
50 7<br />
= − +<br />
2<br />
2x<br />
20x<br />
43<br />
31. Substitute the given points in<strong>to</strong><br />
( 0, 4 ): 4<br />
2<br />
( 0) ( 0)<br />
( 1,0 ): 0 =<br />
2<br />
( 1) + ( 1)<br />
+<br />
( 2,0 ): 0<br />
2<br />
( 2) ( 2)<br />
= a + b + c<br />
a b c<br />
= a + b + c<br />
Simplify these equations:<br />
c = 4 1<br />
.<br />
( )<br />
( )<br />
( )<br />
a+ b+ c = 0 2<br />
4a + 2b + c = 0 3<br />
2<br />
y = ax + bx + c .<br />
Since c = 4, substitute 4 for c in equations (2) and<br />
(3):<br />
a + b + 4 = 0<br />
4a<br />
+ 2b<br />
+ 4 = 0<br />
Simplify these equations:<br />
a + b = −4 4<br />
( )<br />
( )<br />
4a<br />
+ 2b<br />
= −4 5<br />
Eliminate b by multiplying equation (4) by -2 and<br />
add each side <strong>to</strong> the corresponding side of equation<br />
(5):<br />
2a<br />
= 4<br />
a = 2<br />
Next, substitute 2 for a in equation (4):<br />
2 + b = −4<br />
b = −6<br />
Therefore, a = 2, b = -6, and c = 4. So, the equation<br />
2<br />
is y = 2x − 6x<br />
+ 4 .<br />
33. Linear: y = 2x+2 since the slope is 2 and the y-<br />
intercept is (0l, 2);<br />
<strong>Quadratic</strong>: answers may vary. Example:<br />
2<br />
y = 2x<br />
+ 2 ;<br />
Exponential:<br />
y = 2(2) x .<br />
35. Three possible points are (2, 8), (3, 4), and (6, 4).<br />
2<br />
Substitute the given points in<strong>to</strong> y = ax + bx + c .<br />
( 2,8 ): 8<br />
2<br />
( 2) ( 2)<br />
( 3,4 ): 4<br />
2<br />
( 3) ( 3)<br />
( 6, 4 ): 4<br />
2<br />
( 6) ( 6)<br />
= a + b + c<br />
= a + b + c<br />
= a + b + c<br />
Simplify these equations:<br />
4a + 2b + c = 8 1<br />
( )<br />
( )<br />
( )<br />
9a + 3b + c = 4 2<br />
36a + 6b + c = 4 3<br />
Eliminate c by multiplying both sides of equation<br />
(1) by -1:<br />
−4a − 2b − c = − 8 4<br />
( )<br />
Adding the left sides and right sides of equations<br />
(2) and (4) gives:<br />
5a<br />
+ b = − 4 5<br />
( )<br />
Adding the left sides and right sides of equations<br />
(3) and (4) gives:<br />
32a<br />
+ 4b<br />
= − 4 6<br />
Simplify:<br />
8a<br />
+ b = − 1 7<br />
( )<br />
( )<br />
Eliminate b by multiplying equation (5) by -1 and<br />
add each side <strong>to</strong> the corresponding side of equation<br />
(7):<br />
3a<br />
= 3<br />
a = 1<br />
Next, substitute 1 for a in equation (5):<br />
5 1 + b = −4<br />
( )<br />
b = −9<br />
Then, substitute 1 for a and -9 for b in equation (1):<br />
4a + 2b + c = 8<br />
4(1) + 2( − 9) + c = 8<br />
4 − 18 + c = 8<br />
c = 22<br />
Therefore, a = 1, b = -9, and c = 22. So, the<br />
2<br />
equation is y = x − 9x<br />
+ 22 .<br />
2<br />
37. Solve for a in y = a( x − h)<br />
+ k by substituting 5<br />
for h and 8 for k since (5, 8) is the vertex, and<br />
substitute 4 for x and 6 for y since (4, 6) lies on the<br />
parabola:<br />
2<br />
y = a( x − h)<br />
+ k<br />
= − +<br />
2<br />
6 a(4 5) 8<br />
6 = a + 8<br />
a = −2<br />
207
Homework 7.5<br />
SSM: Intermediate Algebra<br />
2<br />
Since a = -2, y = −2( x − 5) + 8 or expanding the<br />
solution:<br />
y = −2( x − 5)( x − 5) − 42<br />
= − − + −<br />
2<br />
2( x 10x<br />
25) 42<br />
= − + − −<br />
2<br />
2x<br />
20x<br />
50 42<br />
2<br />
2x<br />
20x<br />
42<br />
= − + −<br />
.<br />
The quadratic regression equation, which is<br />
2<br />
f ( t) = − 0.66t + 21.49t<br />
− 95.85 , fits the data well.<br />
This equation can be verified by graphing it with<br />
the scattergram:<br />
39. Answers may vary. Example:<br />
( 0, 2 ) , ( 1,1 ), and 6, 4 .<br />
The equation of the parabola through these points<br />
2<br />
is y = x − 2x<br />
+ 2 .<br />
7. a. First draw a scattergram of the data.<br />
Homework 7.5<br />
1. a. A quadratic function would be reasonable.<br />
b. A linear function would be reasonable.<br />
c. An exponential function would be reasonable.<br />
3.<br />
d. None of the mentioned types of functions<br />
would be reasonable for this scattergram.<br />
The linear regression equation is :<br />
f ( t) = 5.09t<br />
+ 11.75<br />
The exponential regression equation is :<br />
f ( t ) = 16.63(1.14) t<br />
The quadratic regression equation is :<br />
2<br />
f ( t) = 0.277t + 1.976t<br />
+ 16.493<br />
b. The exponential model gives the best estimate.<br />
The data does not suggest a quadratic relationship<br />
based on the scattergram above. A quadratic<br />
function is not a reasonable function.<br />
5. First, draw a scattergram of the data.<br />
9. First draw a scattergram of the data:<br />
208
SSM: Intermediate Algebra Homework 7.5<br />
The quadratic regression equation, which is<br />
2<br />
f ( t) = 0.013t − 1.188t<br />
+ 28.239 , fits the data well.<br />
This equation can be verified by graphing it with<br />
the scattergram:<br />
The quadratic function is the best model since<br />
the value of f(d) will eventually return <strong>to</strong> 0 as<br />
the horizontal distance increases.<br />
15. a. First draw a scattergram for C. The quadratic<br />
regression equation,<br />
2<br />
C( t) = 0.90t − 19.94t<br />
+ 121.38 , is the best<br />
model for this data (as can be seen from the<br />
scattergram below.<br />
11. First draw a scattergram of the data:<br />
The quadratic regression equation is, which is<br />
2<br />
f ( t) = 0.0067t − 0.12t<br />
+ 6.39 , fits the data well.<br />
This equation can be verified by graphing it with<br />
the scattergram:.<br />
Next, draw a scattergram of the data for H.<br />
The quadratic regression equation,<br />
H ( t) = 2.48t<br />
− 1.30 , is the best model for this<br />
data (as can be seen from the scattergram<br />
below.<br />
13. a., b. The quadratic function<br />
2<br />
f ( d) = − 0.000333d + 0.152d<br />
+ 4 , is the best<br />
model for the three points (0, 4), (60.5, 12) and<br />
(127.3, 18).<br />
b. Graph the two regression equations on the<br />
same coordinate axes on your calcula<strong>to</strong>r. Use<br />
your calcula<strong>to</strong>r <strong>to</strong> find the points of<br />
intersection.<br />
209
Homework 7.6<br />
SSM: Intermediate Algebra<br />
The intersection points for C and H are<br />
16.8,40.5 . These points<br />
( 8.1,18.8 ) and ( )<br />
mean that the sales of Corona and Heineken<br />
beers were equal in 1988 and in 1997, when<br />
the sales were at 18.8 million cases, and 40.5<br />
million cases, respectively for each company.<br />
b. The parabola has a minimum point at the<br />
vertex, since a is positive. <strong>Using</strong> a graphing<br />
calcula<strong>to</strong>r, find that this minimum point is<br />
approximately (45.77, 1.01).<br />
Homework 7.6<br />
1. a. Solve for t when f ( t ) = 0 :<br />
2<br />
0 0.66t<br />
21.49t<br />
95.85<br />
= − + −<br />
2<br />
21.49 (21.49) 4( 0.66)( 95.85)<br />
− ± − − −<br />
t =<br />
2( −0.66)<br />
t ≈ 5.33 or t ≈ 27.23<br />
The t-intercepts are ( 5.33,0 ) and ( )<br />
27.23,0 .<br />
So, according <strong>to</strong> this model, no firms<br />
performed drug tests on employees in 1985 or<br />
2027.<br />
So, according <strong>to</strong> this model, a 46-year-old<br />
driver has 1 fatal crash per 100 million miles.<br />
5. a. <strong>Using</strong> a graphing calcula<strong>to</strong>r, find that the<br />
maximum point of the parabola is<br />
approximately (228, 21.3).<br />
b. Any value of t that cause the percentage of<br />
firms performing drug tests <strong>to</strong> be negative.<br />
This would occur in years before 1985<br />
5.33<br />
t < 27.23 .<br />
( t < ) and years after 2027 ( )<br />
c. Find the vertex of f:<br />
The t-coordinate of the vertex is<br />
5.33 + 27.23<br />
= 16.28 . Now, find f(16.28):<br />
2<br />
f (16.28)<br />
2<br />
0.66(16.28) 21.49(16.28) 95.85<br />
= − + −<br />
= 79.08<br />
So, the percentage of firms performing drug<br />
tests will reach a maximum of 79% in the year<br />
1996.<br />
3. a. Answers may vary. Example:<br />
So the maximum height is approximately 21.3<br />
feet.<br />
b. Find f(400):<br />
f (400)<br />
2<br />
0.000333(400) 0.152(400) 20.18<br />
= − + +<br />
= 11.52<br />
According <strong>to</strong> the model, the height of the ball<br />
is approximately 11.5 feet when the ball goes a<br />
horizontal distance of 400 feet. Therefore, the<br />
player did hit a home run.<br />
For a person who is 21 years old, find f(21):<br />
7. a.<br />
2<br />
f (30) = − 0.108(30) + 4.29(30) + 21.83<br />
f (21)<br />
2<br />
0.013(21) 1.19(21) 28.24<br />
= − +<br />
= 8.98<br />
According <strong>to</strong> the model, a 21-year-old driver<br />
has approximately 9 fatal crashes per 100<br />
million miles.<br />
= 53.33<br />
According <strong>to</strong> this model, in 2010, 53.33% of<br />
households will have cable. <strong>Model</strong> breakdown<br />
has likely occurred.<br />
b. Solve for t when f ( t ) = 30<br />
210
SSM: Intermediate Algebra Homework 7.6<br />
2<br />
30 0.108t<br />
4.29t<br />
21.83<br />
= − + +<br />
= − + −<br />
2<br />
0 0.108t<br />
4.29t<br />
8.17<br />
− ± − − − −<br />
t =<br />
2( −0.108)<br />
2<br />
4.29 ( 4.29) 4( 0.108)( 8.17)<br />
t ≈ 2.01 or t ≈ 37.72<br />
According <strong>to</strong> this model, 30% of households<br />
had cable in 1982 and will have cable in 2018.<br />
<strong>Model</strong> breakdown has likely occurred for the<br />
2018 prediction.<br />
c. <strong>Using</strong> a graphing calcula<strong>to</strong>r, find that the<br />
maximum point of the parabola is<br />
approximately (19.9, 64.4).<br />
c.<br />
2<br />
0.0067t<br />
0.12t<br />
6.39 300<br />
2<br />
0.0067t<br />
0.12 293.61 0<br />
t =<br />
− + =<br />
− − =<br />
± − −<br />
2<br />
0.12 (0.12) 4(0.0067)( 293.61)<br />
2(0.0067)<br />
t ≈ −200.6 or t ≈ 218.5<br />
So, according <strong>to</strong> the model, the population of<br />
the U.S. was 300 million in 1589 (model<br />
breakdown) and will be 300 million in 2009.<br />
So, according <strong>to</strong> the model, the maximum<br />
percentage of households with cable television<br />
will be 64.4% and this will occur in 2000.<br />
d. The vertex of the parabola is approximately<br />
8.96,5.85 . It can be found by graphing the<br />
( )<br />
model with a graphing calcula<strong>to</strong>r.<br />
d. Since the percent will never go below 0, model<br />
breakdown occurs for years before 1976. Since<br />
the percent decreases for all values of t greater<br />
than 19.9, model breakdown also occurs for<br />
years after 2000.<br />
9. Solve for t when C(t) = H(t)<br />
− + = −<br />
2<br />
0.90t 19.94t 121.38 2.48t<br />
1.30<br />
− + =<br />
2<br />
0.90t<br />
22.42t<br />
122.68 0<br />
t =<br />
± −<br />
2<br />
22.42 (22.42) 4(0.90)(122.68)<br />
t ≈ 8.1 or t ≈ 16.8<br />
2(0.90)<br />
So, sales of Corona and Heineken were about the<br />
same in 1988 and 1997.<br />
e.<br />
The portion of the parabola <strong>to</strong> the left of the<br />
vertex suggests that the population decreased<br />
before 1796, which is false. <strong>Model</strong> breakdown<br />
occurs for years prior <strong>to</strong> 1796.<br />
P<br />
11. a.<br />
2<br />
f (217) = 0.0067(217) − 0.12(217) + 6.39<br />
≈ 295.85<br />
t<br />
b. Let f ( t ) = 300<br />
211
Homework 7.6<br />
SSM: Intermediate Algebra<br />
14. a.<br />
b.<br />
212<br />
E (209) = 3.9(1.030) ≈ 2053.8<br />
According <strong>to</strong> the model E, the U.S. population<br />
was 2.1 billion people in 2002. This is a much<br />
higher [prediction than the actual value of<br />
286.7 million.<br />
= − + −<br />
2<br />
0 1.30t<br />
28.99t<br />
61.29<br />
− ± − − −<br />
t =<br />
2( −1.30)<br />
2<br />
28.99 (28.99) 4( 1.30)( 61.29)<br />
t ≈ 2.36 or t ≈ 19.94<br />
So, according <strong>to</strong> the model, there were no<br />
schools with Internet access in 1992 and there<br />
will be no schools with Internet access in<br />
2010 (model breakdown for the 2010<br />
prediction).<br />
c. f(t) is quadratic, so find the vertex. The vertex<br />
is approximately (11.5, 100.33). It can be<br />
found by graphing the model with a graphing<br />
calcula<strong>to</strong>r.<br />
Yes, there is a great difference in these<br />
predictions. The model E gives more accurate<br />
predictions for years between 1790 and 1860<br />
only.<br />
15. a. First, draw a scattergram of the data.<br />
This means that the model predicts that the<br />
percentage of schools with Internet access will<br />
reach a maximum of 100% in 2001.<br />
2<br />
= ( ) = − 1.30 + 28.99 − 61.29 , is the<br />
p f t t t<br />
best model for this data (as can be seen from<br />
the scattergram below.<br />
d. From part b, we know that the model predicts<br />
there will be no Internet access at 1992 ( t =<br />
2.36) or 2010 (t = 19.94) (the latter is likely<br />
model breakdown). <strong>Model</strong> breakdown occurs<br />
for certain before 1992 (t < 2.36) and after<br />
2010 (t > 19.94) as the percent of schools with<br />
Internet access can not be negative.<br />
e.<br />
Better model<br />
b. Solve for t when f ( t ) = 0 :<br />
212
SSM: Intermediate Algebra<br />
<strong>Chapter</strong> 7 Review Exercises<br />
17. a. Linear: f ( t) = 1.86t<br />
− 1.61<br />
Exponential: f ( t ) = 1.63(1.26) t<br />
<strong>Quadratic</strong>:<br />
f t = t − t +<br />
2<br />
( ) 0.25 0.92 2.62<br />
So, the quadratic model predicts that there will<br />
be enough wind energy <strong>to</strong> meet the needs of<br />
the world in 2049.<br />
e. Answers may vary. Example:<br />
An exponential model gets larger faster than a<br />
quadratic model, so the exponential function<br />
will take less time <strong>to</strong> obtain enough power <strong>to</strong><br />
supply the world.<br />
The quadratic and exponential models fit the<br />
data well.<br />
b. The exponential model, as it gives values<br />
lower than 1.9 (the 1990 capacity) for years<br />
before 1990, whereas the quadratic model<br />
gives values higher than 1.9 for years before<br />
1990.<br />
19. Answers may vary.<br />
<strong>Chapter</strong> 7 Review Exercises<br />
1. 72 = 36⋅ 2 = 36 2 = 6 2<br />
2. 3 = 3 = 3 ⋅ 5 =<br />
15<br />
5 5 5 5 5<br />
c. Since 1 MW meets the needs of 1000 people.<br />
1 thousand MW meets the needs of<br />
(1000)(1000) = 1 million people. There are 6.2<br />
billion people in the world (or, 6200 million<br />
people). So, 6200 thousand MW are needed.<br />
Let f(t) = 6200 and solve for t:<br />
3.<br />
4.<br />
50 50 25⋅<br />
2 5 2<br />
= = =<br />
49 49 7 7<br />
49 49 7<br />
= =<br />
100 100 10<br />
t<br />
6200 = 1.63(1.26)<br />
6200 t<br />
= 1.26<br />
1.63<br />
t ⎛ 6200 ⎞<br />
log(1.26 ) = log⎜ ⎟<br />
⎝ 1.63 ⎠<br />
⎛ 6200 ⎞<br />
t log(1.26) = log⎜ ⎟<br />
⎝ 1.63 ⎠<br />
⎛ 6200 ⎞<br />
log ⎜ ⎟ 1.63<br />
t =<br />
⎝ ⎠<br />
log(1.26)<br />
t ≈ 35.7<br />
So, according <strong>to</strong> the exponential model, there<br />
will be enough wind energy <strong>to</strong> meet the needs<br />
of the world in 2026.<br />
d. Let f(t) = 6200 and solve for t:<br />
2<br />
6200 0.25t<br />
0.92t<br />
2.62<br />
2<br />
0 0.25t<br />
0.92t<br />
6197.38<br />
t =<br />
= − +<br />
= − −<br />
2<br />
0.92 ( 0.92) 4(0.25)( 6197.38)<br />
± − − −<br />
2(0.25)<br />
t ≈ −155.6 (model breakdown) or t ≈ 159.3<br />
5. 6x<br />
2 − 3x<br />
− 2 = 0<br />
− − ± − − −<br />
x =<br />
2(6)<br />
x =<br />
2<br />
( 3) ( 3) 4(6)( 2)<br />
3 ± 57<br />
12<br />
6. 5x 2 = 7<br />
x =<br />
5<br />
2 7<br />
= ±<br />
= ±<br />
7<br />
5<br />
7<br />
5<br />
7 5<br />
= ± ⋅<br />
5 5<br />
= ±<br />
35<br />
5<br />
213
<strong>Chapter</strong> 7 Review Exercises<br />
SSM: Intermediate Algebra<br />
7.<br />
2<br />
5( x − 3) + 4 = 7<br />
2<br />
5( x 3) 3<br />
2<br />
− =<br />
( x 3)<br />
− =<br />
x − 3 = ±<br />
3<br />
5<br />
3<br />
5<br />
3 5<br />
x − 3 = ± ⋅<br />
5 5<br />
x − 3 = ±<br />
x = 3 ±<br />
x =<br />
8. x 2 = 98<br />
x = ±<br />
15<br />
5<br />
15<br />
5<br />
15 ± 15<br />
5<br />
98<br />
= ± 49⋅<br />
2<br />
= ± 7 2<br />
9. ( x + 1)( x − 7) = 4<br />
10.<br />
− 7 + − 7 = 4<br />
2<br />
x x x<br />
2<br />
x − 6x<br />
− 11 = 0<br />
− − ± − − −<br />
x =<br />
2(1)<br />
=<br />
=<br />
2<br />
( 6) ( 6) 4(1)( 11)<br />
6 ± 80<br />
2<br />
6 ± 4 5<br />
2<br />
= 3 ± 2 5<br />
− + =<br />
2<br />
2( x 4) 9<br />
( x + 4) = −<br />
2<br />
2 9<br />
9<br />
x + 4 = ± −<br />
2<br />
Since the square of a negative number is not a real<br />
number, there are no real solutions.<br />
11. 3x<br />
2 = 6x<br />
− =<br />
2<br />
3x<br />
6x<br />
0<br />
3 x( x − 2) = 0<br />
3x<br />
= 0 or x − 2 = 0<br />
x = 0 or x = 2<br />
12. x<br />
2 − 2x<br />
− 24 = 0<br />
( x − 6)( x + 4) = 0<br />
x − 6 = 0 or x + 4 = 0<br />
x = 6 or x = −4<br />
13. 4x 2 − 25 = 0<br />
(2x<br />
− 5)(2x<br />
+ 5) = 0<br />
2x<br />
− 5 = 0 or 2x<br />
+ 5 = 0<br />
2x<br />
= 5 or 2x<br />
= −5<br />
5 5<br />
x = or x = −<br />
2 2<br />
14. 3x<br />
2 − 7x<br />
+ 2 = 0<br />
(3x<br />
−1)( x − 2) = 0<br />
15.<br />
3x<br />
− 1 = 0 or x − 2 = 0<br />
3x<br />
= 1 or x = 2<br />
1<br />
x = or x = 2<br />
3<br />
2<br />
25x − 9 = 0<br />
(5x<br />
− 3)(5x<br />
+ 3) = 0<br />
5x<br />
− 3 = 0 or 5x<br />
+ 3 = 0<br />
5x<br />
= 3 or 5x<br />
= −3<br />
3 3<br />
x = or x = −<br />
5 5<br />
16. 3x 2 − 1 = x 2 − 5x<br />
+ 1<br />
2<br />
2x<br />
5x<br />
2 0<br />
− ± − −<br />
x =<br />
2(2)<br />
x =<br />
+ − =<br />
2<br />
5 5 4(2)( 2)<br />
− 5 ± 41<br />
4<br />
17. 2x<br />
2 = 4 − 5x<br />
2<br />
2x<br />
5x<br />
4 0<br />
+ − =<br />
− ± − −<br />
x =<br />
2(5)<br />
2<br />
5 5 4(2)( 4)<br />
− 5 ± 57<br />
x =<br />
10<br />
214
SSM: Intermediate Algebra<br />
<strong>Chapter</strong> 7 Review Exercises<br />
18.<br />
19.<br />
20.<br />
21.<br />
− =<br />
2<br />
4x<br />
x 1<br />
2<br />
x − 4x<br />
= −1<br />
2<br />
x − 4x<br />
+ 4 = − 1+<br />
4<br />
− =<br />
2<br />
( x 2) 3<br />
x − 2 = ± 3<br />
x = 2 ± 3<br />
− + = − + +<br />
2<br />
2x<br />
− 5 = 0<br />
2 2<br />
3x x 5x x 5<br />
2<br />
2x<br />
5<br />
x<br />
2<br />
=<br />
=<br />
x = ±<br />
x = ±<br />
5<br />
2<br />
5<br />
2<br />
5<br />
2<br />
5 2<br />
x = ± ⋅<br />
2 2<br />
x = ±<br />
10<br />
2<br />
− + = +<br />
2<br />
2 x(2x 5) 13 3x<br />
5<br />
2 2<br />
4x 10x 13 3x<br />
5<br />
2<br />
x − 10x<br />
+ 8 = 0<br />
− − ± − −<br />
x =<br />
2(1)<br />
x =<br />
x =<br />
− + = +<br />
2<br />
( 10) ( 10) 4(1)(8)<br />
10 ± 68<br />
2<br />
10 ± 2 17<br />
2<br />
x = 5 ± 17<br />
2<br />
3 x( x 7) 13 2x<br />
7<br />
− + = −<br />
− + = −<br />
2 2<br />
3x 21x 13 2x<br />
7<br />
2<br />
x − 21x<br />
+ 20 = 0<br />
( x − 20)( x − 1) = 0<br />
x − 20 = 0 or x − 1 = 0<br />
x = 20 or x = 1<br />
22.<br />
23.<br />
24.<br />
25.<br />
2 2<br />
2( x x) 3( x 2 x) 5<br />
− = − +<br />
2 2<br />
2x 2x 3x<br />
6 5<br />
2<br />
x − 4x<br />
+ 5 = 0<br />
− − ± − −<br />
x =<br />
2(1)<br />
x =<br />
− = − +<br />
2<br />
( 4) ( 4) 4(1)(5)<br />
4 ± −4<br />
2<br />
Since the square of a negative number is not a real<br />
number, there are no real solutions.<br />
2 2<br />
( x 2) ( x 3) 2<br />
+ + − =<br />
+ 4 + 4 + − 6 + 9 = 2<br />
2 2<br />
x x x x<br />
2<br />
2x<br />
2x<br />
13 2<br />
2<br />
2x<br />
2x<br />
11 0<br />
2<br />
( 2) ( 2) 4(2)(11)<br />
− − ± − −<br />
x =<br />
2(2)<br />
x =<br />
− + =<br />
− + =<br />
2 ± −84<br />
4<br />
Since the square of a negative number is not a real<br />
number, there are no real solutions.<br />
2<br />
5(5x − 8) = 0<br />
2<br />
25x<br />
40 9<br />
2<br />
25x<br />
49<br />
x<br />
2<br />
=<br />
x = ±<br />
x = ±<br />
− =<br />
=<br />
49<br />
25<br />
49<br />
25<br />
7<br />
5<br />
− =<br />
2<br />
2.7x<br />
5.1x<br />
9.4<br />
− − =<br />
2<br />
2.7x<br />
5.1x<br />
9.4 0<br />
− − ± − − −<br />
x =<br />
2(2.7)<br />
2<br />
( 5.1) ( 5.1) 4(2.7)( 9.4)<br />
x ≈ −1.15 or x ≈ 3.04<br />
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<strong>Chapter</strong> 7 Review Exercises<br />
SSM: Intermediate Algebra<br />
26.<br />
2<br />
1.7( x 2.3) 3.4 2.8<br />
− = −<br />
2<br />
1.7x<br />
3.91 3.4 2.8<br />
− = −<br />
+ − =<br />
2<br />
1.7x<br />
2.8x<br />
7.31 0<br />
− ± − −<br />
x =<br />
2(1.7)<br />
2<br />
2.8 2.8 4(1.7)( 7.31)<br />
x ≈ −3.05 or x ≈ 1.41<br />
27. x<br />
2 + 6x<br />
− 4 = 0<br />
2<br />
x + 6x<br />
= 4<br />
2<br />
x + 6x<br />
+ 9 = 4 + 9<br />
+ =<br />
2<br />
( x 3) 13<br />
x + 3 = ± 13<br />
x = − 3 ± 13<br />
28. 8x<br />
2 + 12x<br />
+ 8 = 0<br />
2 3<br />
x + x + 1 = 0<br />
2<br />
2 3 9 9<br />
x + x + = − 1+<br />
2 16 16<br />
2<br />
⎛ 3 ⎞ 7<br />
⎜ x + ⎟ = −<br />
⎝ 4 ⎠ 16<br />
3 7<br />
x + = ± −<br />
4 16<br />
x<br />
x<br />
Since the square of a negative number is not a real<br />
number, there are no real solutions.<br />
29. 2x<br />
2 = − 3x<br />
+ 6<br />
2<br />
2x<br />
+ 3x<br />
= 6<br />
2 3<br />
x + x = 3<br />
2<br />
2 3 9 9<br />
x + x + = 3 +<br />
2 16 16<br />
⎛ 3 ⎞ 57<br />
⎜ x + ⎟ =<br />
⎝ 4 ⎠ 16<br />
3 57<br />
x + = ±<br />
4 16<br />
3 57<br />
x + = ±<br />
4 4<br />
3 57<br />
x = − ±<br />
4 4<br />
x =<br />
2<br />
− 3 ± 57<br />
4<br />
30. x( x − 3) + 2 = 2 x( x + 1) + 1<br />
− 3 + 2 = 2 + 2 + 1<br />
2 2<br />
x x x x<br />
2<br />
1 x 5<br />
2 25 25<br />
x + 5x<br />
+ = 1+<br />
4 4<br />
⎛ 5 ⎞ 29<br />
⎜ x + ⎟ =<br />
⎝ 2 ⎠ 4<br />
2<br />
5 29<br />
x + = ±<br />
2 4<br />
5 29<br />
x + = ±<br />
2 2<br />
x<br />
5 29<br />
x = − ±<br />
2 2<br />
x =<br />
= +<br />
− 5 ± 29<br />
2<br />
31. Solve for x when f ( x ) = 0 :<br />
2<br />
3x<br />
10 0<br />
2<br />
3x<br />
10<br />
x<br />
2<br />
=<br />
x = ±<br />
10<br />
3<br />
10<br />
3<br />
10 3<br />
x = ± ⋅<br />
3 3<br />
x = ±<br />
− =<br />
=<br />
30<br />
3<br />
⎛<br />
The x-intercepts are<br />
⎜<br />
⎝<br />
30 ,0<br />
3<br />
32. Solve for x when g( x ) = 0 :<br />
− + + =<br />
2<br />
2( x 3) 5 0<br />
− + = −<br />
2<br />
2( x 3) 5<br />
2<br />
+ =<br />
( x 3)<br />
x + 3 = ±<br />
5<br />
2<br />
5<br />
2<br />
⎞<br />
⎟<br />
⎠<br />
and<br />
⎛<br />
⎜<br />
−<br />
⎝<br />
30 ,0<br />
3<br />
⎞<br />
⎟<br />
.<br />
⎠<br />
216
SSM: Intermediate Algebra<br />
<strong>Chapter</strong> 7 Review Exercises<br />
5 2<br />
x + 3 = ± ⋅<br />
2 2<br />
x + 3 = ±<br />
x = − 3 ±<br />
x =<br />
10<br />
2<br />
10<br />
2<br />
− 6 ± 10<br />
2<br />
⎛ − 6 + 10 ⎞<br />
The x-intercepts are<br />
⎜ ,0<br />
2 ⎟<br />
and<br />
⎝ ⎠<br />
⎛ −6 − 10 ⎞<br />
⎜ ,0<br />
2 ⎟<br />
.<br />
⎝ ⎠<br />
33. Solve for x when h( x ) = 0 :<br />
2<br />
3x<br />
2x<br />
2 0<br />
− ± − −<br />
x =<br />
2(3)<br />
x =<br />
x =<br />
x =<br />
+ − =<br />
2<br />
2 2 4(3)( 2)<br />
− 2 ± 28<br />
6<br />
− 2 ± 2 7<br />
6<br />
− 1±<br />
7<br />
3<br />
⎛ −1−<br />
7 ⎞<br />
The x-intercepts are<br />
⎜ ,0<br />
3 ⎟<br />
and<br />
⎝ ⎠<br />
⎛ − 1+<br />
7 ⎞<br />
⎜ ,0<br />
3 ⎟<br />
.<br />
⎝ ⎠<br />
34. Solve for x when k( x ) = 0 :<br />
− + − =<br />
2<br />
5x<br />
3x<br />
1 0<br />
− ± − − −<br />
x =<br />
2( −5)<br />
2<br />
3 3 4( 5)( 1)<br />
− 3 ± −11<br />
x =<br />
−10<br />
Since the square root of a negative number is not a<br />
real number, there are no real number solutions.<br />
Therefore, there are no x-intercepts.<br />
35. Solve for x when f ( x ) = 0 :<br />
2<br />
2x<br />
3x<br />
4 0<br />
− − ± − − −<br />
x =<br />
2(2)<br />
x =<br />
− − =<br />
2<br />
( 3) ( 3) 4(2)( 4)<br />
3 ± 41<br />
4<br />
⎛ 3 + 41 ⎞<br />
The x-intercepts are<br />
,0<br />
and<br />
⎜ 4 ⎟<br />
⎝ ⎠<br />
⎛ 3 − 41 ⎞<br />
,0<br />
.<br />
⎜ 4 ⎟<br />
⎝ ⎠<br />
36. Solve for x when g( x ) = 0 :<br />
− + + =<br />
2<br />
3x<br />
5x<br />
7 0<br />
− ± − −<br />
x =<br />
2( −3)<br />
2<br />
5 5 4( 3)(7)<br />
− 5 ± 109<br />
x =<br />
−6<br />
− 1(5 ± 109)<br />
x =<br />
−6<br />
x =<br />
5 ± 109<br />
6<br />
⎛ 5 + 109 ⎞<br />
The x-intercepts are<br />
⎜ ,0<br />
6 ⎟<br />
and<br />
⎝ ⎠<br />
⎛ 5 − 109 ⎞<br />
⎜ ,0<br />
6 ⎟<br />
.<br />
⎝ ⎠<br />
37. Fac<strong>to</strong>ring:<br />
2<br />
x − 2x<br />
− 8 = 0<br />
( x − 4)( x + 2) = 0<br />
x − 4 = 0 or x + 2 = 0<br />
x = 4 or x = −2<br />
Completing the square:<br />
2<br />
x − 2x<br />
− 8 = 0<br />
2<br />
x − 2x<br />
= 8<br />
2<br />
x − 2x<br />
+ 1 = 8 + 1<br />
− =<br />
2<br />
( x 1) 9<br />
x − 1 = ± 9<br />
x − 1= ± 3<br />
x − 1 = 3 or x − 1= −3<br />
x = 4 or x = −2<br />
217
<strong>Chapter</strong> 7 Review Exercises<br />
SSM: Intermediate Algebra<br />
<strong>Quadratic</strong> Formula:<br />
2<br />
x − 2x<br />
− 8 = 0<br />
− − ± − − −<br />
x =<br />
2(1)<br />
2<br />
( 2) ( 2) 4(1)( 8)<br />
2 ± 36<br />
x =<br />
2<br />
2 ± 6<br />
x =<br />
2<br />
2 + 6 8 2 − 6 −4<br />
x = = or x = =<br />
2 2 2 2<br />
x = 4 or x = −2<br />
38. Find k when the discriminant equals 0:<br />
2<br />
b − 4ac<br />
= 0<br />
k<br />
k<br />
k<br />
39. a.<br />
2<br />
2<br />
2<br />
− 4(3)(12) = 0<br />
− 144 = 0<br />
= 144<br />
x = ±<br />
x = ± 12<br />
b.<br />
144<br />
− + =<br />
2<br />
3x<br />
6x<br />
7 3<br />
2<br />
3x<br />
6x<br />
4 0<br />
− − ± − −<br />
x =<br />
2(3)<br />
x =<br />
− + =<br />
2<br />
( 6) ( 6) 4(3)(4)<br />
6 ± −12<br />
6<br />
Since the square root of a negative is not a real<br />
number, there are no real number solutions.<br />
There is no such value for x.<br />
− + =<br />
2<br />
3x<br />
− 6x<br />
+ 3 = 0<br />
2<br />
3x<br />
6x<br />
7 4<br />
− − ± − −<br />
x =<br />
2(3)<br />
x<br />
x<br />
2<br />
( 6) ( 6) 4(3)(3)<br />
6 ± 0<br />
=<br />
6<br />
6<br />
= = 1<br />
6<br />
c.<br />
2<br />
3x<br />
− 6x<br />
+ 7 = 5<br />
2<br />
3x<br />
− 6x<br />
+ 2 = 0<br />
− − ± − −<br />
x =<br />
2(3)<br />
x =<br />
x =<br />
x =<br />
2<br />
( 6) ( 6) 4(3)(2)<br />
6 ± 12<br />
6<br />
6 ± 2 3<br />
6<br />
3 ± 3<br />
3<br />
d. Answers may vary.<br />
2<br />
40. <strong>Using</strong> the equation y = a( x − h)<br />
+ k , substitute -4<br />
for h and 3 for k since the vertex is (-4, 3). Also,<br />
substitute -3 for x and 6 for y since (-3, 6) lies on<br />
the parabola.<br />
2<br />
y = a( x − h)<br />
+ k<br />
2<br />
6 a( 3 ( 4)) 3<br />
= − − − +<br />
= − + +<br />
2<br />
6 a( 3 4) 3<br />
6 = a(1) + 3<br />
3 = a<br />
So, the equation is<br />
y = + +<br />
2<br />
3( x 4) 3<br />
41. Substitute the given points in<strong>to</strong><br />
2<br />
( 1,3 ): 3 = a( 1) + b( 1)<br />
+ c<br />
( 2,6 ): 6 =<br />
2<br />
( 2) + ( 2)<br />
+<br />
( 3,13 ):13 2<br />
( 3) ( 3)<br />
a b c<br />
= a + b + c<br />
Simplify these equations:<br />
a+ b+ c = 3 1<br />
( )<br />
( )<br />
( )<br />
4a + 2b + c = 6 2<br />
9a + 3b + c = 13 3<br />
2<br />
y = ax + bx + c .<br />
Eliminate c by multiplying both sides of equation<br />
(1) by -1:<br />
−a − b − c = − 3 4<br />
( )<br />
Adding the left sides and right sides of equations<br />
(2) and (4) gives:<br />
3a<br />
+ b = 3 5<br />
( )<br />
Adding the left sides and right sides of equations<br />
(3) and (4) gives:<br />
8a<br />
+ 2b<br />
= 10 6<br />
Simplify:<br />
4a<br />
+ b = 5 7<br />
( )<br />
( )<br />
218
SSM: Intermediate Algebra<br />
<strong>Chapter</strong> 7 Review Exercises<br />
Eliminate b by multiplying equation (5) by -1 and<br />
add each side <strong>to</strong> the corresponding side of equation<br />
(7):<br />
a = 2<br />
Next, substitute 2 for a in equation (5):<br />
3 2 + b = 3<br />
( )<br />
b = −3<br />
Then, substitute 2 for a and -3 for b in equation (1):<br />
a+ b+ c = 3<br />
2 + ( − 3) + c = 3<br />
c = 4<br />
Therefore, a = 2, b = -3, and c = 4. So, the equation<br />
2<br />
is y = 2x − 3x<br />
+ 4 .<br />
42. Substitute the given points in<strong>to</strong><br />
2<br />
( 1,4 ): 4 = a( 1) + b( 1)<br />
+ c<br />
( 3, −2 ): − 2 =<br />
2<br />
( 3) + ( 3)<br />
+<br />
( 4, 11 ): 11<br />
2<br />
( 4) ( 4)<br />
a b c<br />
− − = a + b + c<br />
Simplify these equations:<br />
a+ b+ c = 4 1<br />
( )<br />
( )<br />
( )<br />
9a + 3b + c = −2 2<br />
16a + 4b + c = −11 3<br />
2<br />
y = ax + bx + c .<br />
Eliminate c by multiplying both sides of equation<br />
(1) by -1:<br />
−a − b − c = − 4 4<br />
( )<br />
Adding the left sides and right sides of equations<br />
(2) and (4) gives:<br />
8a<br />
+ 2b<br />
= − 6 5<br />
Simplify:<br />
4a<br />
+ b = − 3 6<br />
( )<br />
( )<br />
Adding the left sides and right sides of equations<br />
(3) and (4) gives:<br />
15a<br />
+ 3b<br />
= − 15 7<br />
Simplify:<br />
5a<br />
+ b = − 5 8<br />
( )<br />
( )<br />
Eliminate b by multiplying equation (6) by -1 and<br />
add each side <strong>to</strong> the corresponding side of equation<br />
(8):<br />
a = − 2<br />
Next, substitute -2 for a in equation (6):<br />
4 − 2 + b = −3<br />
( )<br />
b = 5<br />
Then, substitute -2 for a and 5 for b in equation (1):<br />
a+ b+ c = 4<br />
− 2 + 5 + c = 4<br />
c = 1<br />
Therefore, a = -2, b = 5, and c = 1. So, the equation<br />
2<br />
is y = − 2x + 5x<br />
+ 1.<br />
43. Substitute the given points in<strong>to</strong><br />
( 2,9 ): 9 =<br />
2<br />
( 2) + ( 2)<br />
+<br />
( 3,18 ):18 2<br />
( 3) ( 3)<br />
( 5,48 ): 48<br />
2<br />
( 5) ( 5)<br />
a b c<br />
= a + b + c<br />
= a + b + c<br />
Simplify these equations:<br />
4a + 2b + c = 9 1<br />
( )<br />
( )<br />
( )<br />
9a + 3b + c = 18 2<br />
25a + 5b + c = 48 3<br />
2<br />
y = ax + bx + c .<br />
Eliminate c by multiplying both sides of equation<br />
(1) by -1:<br />
−4a − 2b − c = − 9 4<br />
( )<br />
Adding the left sides and right sides of equations<br />
(2) and (4) gives:<br />
5a<br />
+ b = 9 5<br />
( )<br />
Adding the left sides and right sides of equations<br />
(3) and (4) gives:<br />
21a<br />
+ 3b<br />
= 39 6<br />
Simplify:<br />
7a<br />
+ b = 13 7<br />
( )<br />
( )<br />
Eliminate b by multiplying equation (5) by -1 and<br />
add each side <strong>to</strong> the corresponding side of equation<br />
(7):<br />
2a<br />
= 4<br />
a = 2<br />
Next, substitute 2 for a in equation (5):<br />
5 2 + b = 9<br />
( )<br />
b = −1<br />
Then, substitute 2 for a and -1 for b in equation (1):<br />
4a + 2b + c = 9<br />
4(2) + 2( − 1) + c = 9<br />
8 − 2 + c = 9<br />
c = 3<br />
Therefore, a = 2, b = -1, and c = 3. So, the equation<br />
2<br />
is y = 2x − x + 3 .<br />
219
<strong>Chapter</strong> 7 Review Exercises<br />
SSM: Intermediate Algebra<br />
44. Substitute the given points in<strong>to</strong><br />
2<br />
( 0,5 ): 5 = a( 0) + b( 0)<br />
+ c<br />
2<br />
( 2,3 ): 3 = a( 2) + b( 2)<br />
+ c<br />
2<br />
( 4, 15 ): 15 ( 4) ( 4)<br />
− − = a + b + c<br />
Simplify these equations:<br />
c = 5 1<br />
( )<br />
( )<br />
( )<br />
4a + 2b + c = 3 2<br />
16a + 4b + c = −15 3<br />
2<br />
y = ax + bx + c .<br />
Since c = 5, substitute 5 for c in equations (2) and<br />
(3):<br />
4a<br />
+ 2 b + (5) = 3<br />
16a<br />
+ 4 b + (5) = −15<br />
Simplifying these equations gives:<br />
4a<br />
+ 2b<br />
= −2 4<br />
( )<br />
( )<br />
16a<br />
+ 4b<br />
= −20 5<br />
To eliminate b, multiply both sides of equation (4)<br />
by -2 and add each side <strong>to</strong> the corresponding side<br />
in equation (5):<br />
8a<br />
= −16<br />
a = −2<br />
Next, substitute -2 for a in equation (4):<br />
4a<br />
+ 2b<br />
= −2<br />
4( − 2) + 2b<br />
= −2<br />
2b<br />
= 6<br />
b = 3<br />
Therefore, a = -2, b = 3, and c = 5. So, the<br />
2<br />
equation is y = − 2x + 3x<br />
+ 5 .<br />
45. Substitute the given points in<strong>to</strong><br />
2<br />
( 0,7 ): 7 = ( 0) + ( 0)<br />
+<br />
2<br />
( 1,8 ): 8 = a( 1) + b( 1)<br />
+ c<br />
2<br />
( 3, 8 ): 8 ( 3) ( 3)<br />
a b c<br />
− − = a + b + c<br />
Simplify these equations:<br />
c = 7 1<br />
( )<br />
( )<br />
( )<br />
a+ b+ c = 8 2<br />
9a + 3b + c = −8 3<br />
2<br />
y = ax + bx + c .<br />
Since c = 7, substitute 7 for c in equations (2) and<br />
(3):<br />
a + b + (7) = 8<br />
9a<br />
+ 3 b + (7) = −8<br />
Simplifying these equations gives:<br />
( )<br />
( )<br />
a + b = 1 4<br />
9a<br />
+ 3b<br />
= −15 5<br />
To eliminate b, multiply both sides of equation (4)<br />
by -3 and add each side <strong>to</strong> the corresponding side<br />
in equation (5):<br />
6a<br />
= −18<br />
a = −3<br />
Next, substitute -3 for a in equation (4):<br />
a + b = 1<br />
( − 3) + b = 1<br />
b = 4<br />
Therefore, a = -3, b = 4, and c = 7. So, the<br />
2<br />
equation is y = − 3x + 4x<br />
+ 7 .<br />
46. Substitute the given points in<strong>to</strong><br />
( 0,3 ): 3 =<br />
2<br />
( 0) + ( 0)<br />
+<br />
( 2,13 ):13 2<br />
( 2) ( 2)<br />
( 3,24 ): 24<br />
2<br />
( 3) ( 3)<br />
a b c<br />
= a + b + c<br />
= a + b + c<br />
Simplify these equations:<br />
c = 3 1<br />
( )<br />
( )<br />
( )<br />
4a + 2b + c = 13 2<br />
9a + 3b + c = 24 3<br />
2<br />
y = ax + bx + c .<br />
Since c = 3, substitute 3 for c in equations (2) and<br />
(3):<br />
4a<br />
+ 2 b + (3) = 13<br />
9a<br />
+ 3 b + (3) = 24<br />
Simplifying these equations gives:<br />
4a<br />
+ 2b<br />
= 10 4<br />
( )<br />
( )<br />
9a<br />
+ 3b<br />
= 21 5<br />
To eliminate b, multiply both sides of equation (4)<br />
by -3 and multiply both sides of equation (5) by 2<br />
−12a<br />
− 6b<br />
= −30 6<br />
( )<br />
( )<br />
18a<br />
+ 6b<br />
= 42 7<br />
To solve for a, add the corresponding sides of<br />
equations (6) and (7) <strong>to</strong>gether:<br />
6a<br />
= 12<br />
a = 2<br />
Next, substitute 2 for a in equation (4):<br />
4a<br />
+ 2b<br />
= 10<br />
4(2) + 2b<br />
= 10<br />
2b<br />
= 2<br />
b = 1<br />
Therefore, a = 2, b = 1, and c = 3. So, the equation<br />
2<br />
is y = 2x + x + 3 .<br />
220
SSM: Intermediate Algebra<br />
<strong>Chapter</strong> 7 Review Exercises<br />
47. Linear:<br />
2 − 4<br />
slope = m = = −2<br />
1−<br />
0<br />
y − intercept = (0,4)<br />
So, y = − 2x<br />
+ 4<br />
Exponential:<br />
As the value of x increases by 1, the value of y is<br />
1<br />
multiplied by ½,, so the base b = . The y-<br />
2<br />
intercept is (0, 4). Therefore, the exponential<br />
x<br />
⎛ 1 ⎞<br />
function is y = 4 ⎜ ⎟<br />
⎝ 2 ⎠<br />
2a<br />
+ b = −3<br />
2(1) + b = −3<br />
b = −5<br />
Therefore, a = 1, b = -5, and c = 7. So, the<br />
2<br />
equation is y = x − 5x<br />
+ 7 .<br />
49. a. First, draw a scattergram of the data. The<br />
quadratic regression equation,<br />
2<br />
f ( t) = − 0.0123t + 1.85t<br />
+ 0.64 , is the best<br />
model for this data.<br />
<strong>Quadratic</strong>:<br />
Answers may vary. Example:<br />
48. Substitute the given points in<strong>to</strong><br />
( 0,7 ): 7<br />
2<br />
( 0) ( 0)<br />
( 2,1 ):1 2<br />
( 2) ( 2)<br />
( 5,7 ): 7<br />
2<br />
( 5) ( 5)<br />
= a + b + c<br />
= a + b + c<br />
= a + b + c<br />
Simplify these equations:<br />
c = 7 1<br />
( )<br />
( )<br />
( )<br />
4a + 2b + c = 1 2<br />
25a + 5b + c = 7 3<br />
y = − +<br />
2<br />
2x<br />
4<br />
2<br />
y = ax + bx + c .<br />
Since c = 7, substitute 7 for c in equations (2) and<br />
(3):<br />
4a<br />
+ 2 b + (7) = 1<br />
25a<br />
+ 5 b + (7) = 7<br />
Simplifying these equations gives:<br />
4a<br />
+ 2b<br />
= −6 4<br />
( )<br />
( )<br />
25a<br />
+ 5b<br />
= 0 5<br />
Simplify these equations:<br />
2a<br />
+ b = −3 6<br />
( )<br />
( )<br />
5a<br />
+ b = 0 7<br />
To eliminate b, multiply both sides of equation (6)<br />
by -1 and add each side <strong>to</strong> the corresponding side<br />
in equation (7):<br />
3a<br />
= 3<br />
a = 1<br />
Next, substitute 1 for a in equation (5):<br />
b. Find f(18)<br />
2<br />
f (18) = − 0.0123(18) + 1.85(19) + 0.64 ≈ 29.9<br />
So, about 30% of 18-year-old Americans<br />
voted.<br />
c. Solve for t when f(t) = 50<br />
= − + +<br />
2<br />
50 0.0123t<br />
1.85t<br />
0.64<br />
= − + −<br />
2<br />
0 0.0123t<br />
1.85t<br />
49.36<br />
− ± − − −<br />
t =<br />
2( −0.0123)<br />
2<br />
1.85 1.85 4( 0.0123)( 49.36)<br />
t ≈ 34.7 or t ≈ 115.7<br />
t ≈ 115.7 is model breakdown. So, according<br />
<strong>to</strong> the model, 50% of people at age 35 vote.<br />
d. <strong>Using</strong> a graphing calcula<strong>to</strong>r, find that the<br />
maximum point of the parabola is<br />
approximately (75.3, 70.2).<br />
So the age of Americans who are most likely<br />
<strong>to</strong> vote is age 75.<br />
221
<strong>Chapter</strong> 7 Test<br />
SSM: Intermediate Algebra<br />
50. a.<br />
e. Answers may vary. Example:<br />
Candidates tend <strong>to</strong> focus on issues for older<br />
people because they are the most likely <strong>to</strong><br />
vote.<br />
b. <strong>Using</strong> quadratic regression:<br />
2<br />
f ( t) = 0.037t − 0.47t<br />
+ 31.84<br />
c.<br />
d.<br />
2<br />
f (37) = 0.037(37) − 0.47(37) + 31.84 ≈ 65.1<br />
This means that in 2001, 65.1% of the grades<br />
at Prince<strong>to</strong>n will be A’s.<br />
= − +<br />
2<br />
100 0.037t<br />
0.47t<br />
31.84<br />
2<br />
0 0.037t<br />
0.47t<br />
68.16<br />
t =<br />
<strong>Chapter</strong> 7 Test<br />
= − −<br />
2<br />
0.47 ( 0.47) 4(0.037)( 68.16)<br />
± − − −<br />
2(0.037)<br />
t ≈ −37.04 or t ≈ 49.74<br />
In 1933 and in 20020, all grades at Prince<strong>to</strong>n<br />
will be A’s. <strong>Model</strong> breakdown has occurred.<br />
1. 32 = 16 ⋅ 2 = 16 2 = 4 2<br />
2. 7 = 7 ⋅ 2 =<br />
7 2<br />
2 2 2 2<br />
3.<br />
20 20 2 5 2 5 3 2 15<br />
= = = ⋅ =<br />
75 75 5 3 5 3 3 15<br />
4. x<br />
2 − 3x<br />
− 10 = 0<br />
( x − 5)( x + 2) = 0<br />
x − 5 = 0 or x + 2 = 0<br />
x = 5 or x = −2<br />
5. 6x 2 = 100<br />
2 100<br />
x =<br />
6<br />
2 50<br />
x =<br />
3<br />
6.<br />
7.<br />
x = ±<br />
x = ±<br />
50<br />
3<br />
50<br />
3<br />
5 2 3<br />
x = ± ⋅<br />
3 3<br />
x = ±<br />
5 6<br />
3<br />
− + − =<br />
2<br />
3x<br />
x 4 0<br />
− ± − − −<br />
x =<br />
2( −3)<br />
2<br />
1 1 4( 3)( 4)<br />
1± −47<br />
=<br />
−6<br />
Since the square of a negative number is not a real<br />
number, there are no real solutions.<br />
2<br />
4( x − 3) + 1 = 7<br />
2<br />
4( x 3) 6<br />
2<br />
− =<br />
( x 3)<br />
2<br />
− =<br />
( x 3)<br />
− =<br />
x − 3 = ±<br />
6<br />
4<br />
3<br />
2<br />
3<br />
2<br />
3 2<br />
x − 3 = ± ⋅<br />
2 2<br />
x − 3 = ±<br />
x = 3 ±<br />
6<br />
2<br />
6<br />
2<br />
6 ± 6<br />
x =<br />
2<br />
8. 3x<br />
2 − 21x<br />
= 0<br />
3 x( x − 7) = 0<br />
3x<br />
= 0 or x − 7 = 0<br />
x = 0 or x = 7<br />
222
SSM: Intermediate Algebra<br />
<strong>Chapter</strong> 7 Test<br />
9. x 2 − 81 = 0<br />
( x − 9)( x + 9) = 0<br />
x − 9 = 0 or x + 9 = 0<br />
x = 9 or x = −9<br />
10. ( x − 3)( x + 5) = 6<br />
2<br />
x x x<br />
+ 5 − 3 − 15 = 6<br />
2<br />
x + 2x<br />
− 21 = 0<br />
2<br />
x + 2x<br />
= 21<br />
2<br />
x + 2x<br />
+ 1 = 21+<br />
1<br />
2<br />
( x 1) 22<br />
+ =<br />
x + 1 = ± 22<br />
x = − 1±<br />
22<br />
14.<br />
2<br />
3.7x<br />
2.4 5.9<br />
= −<br />
2<br />
3.7x<br />
5.9x<br />
2.4 0<br />
x<br />
+ − =<br />
2<br />
5.9 5.9 4(3.7)( 2.4)<br />
− ± − −<br />
x =<br />
2(3.7)<br />
x ≈ −1.93 or x ≈ 0.34<br />
15. x<br />
2 − 8x<br />
− 2 = 0<br />
2<br />
x − 8x<br />
= 2<br />
2<br />
x − 8x<br />
+ 16 = 2 + 16<br />
2<br />
( x 4) 18<br />
− =<br />
x − 4 = ± 18<br />
x − 4 = ± 3 2<br />
x = 4 ± 3 2<br />
11.<br />
2<br />
5( x 4) 10<br />
− + =<br />
2<br />
( x 4) 2<br />
+ = −<br />
x + 4 = ± −2<br />
Since the square of a negative number is not a real<br />
number, there are no real solutions.<br />
12. 2 x( x + 5) = 4x<br />
− 3<br />
2<br />
2x 10x 4x<br />
3<br />
+ = −<br />
2<br />
2x<br />
6x<br />
3 0<br />
+ + =<br />
− ±<br />
x =<br />
x =<br />
x =<br />
x =<br />
2<br />
6 6 4(2)(3)<br />
−<br />
2(2)<br />
− 6 ± 12<br />
4<br />
− 6 ± 2 3<br />
4<br />
− 3 ± 3<br />
2<br />
13. 9x<br />
2 = 16 + 24x<br />
2<br />
9x<br />
− 24x<br />
− 16 = 0<br />
2<br />
( 24) ( 24) 4(9)( 16)<br />
− − ± − − −<br />
x =<br />
2(9)<br />
24 ± 1152<br />
x =<br />
18<br />
x =<br />
24 ± 24 2<br />
18<br />
4 ± 4 3<br />
x =<br />
3<br />
16.<br />
2<br />
2( x 4) 3<br />
− = −<br />
2<br />
2x<br />
8 3<br />
−<br />
= −<br />
2<br />
2x<br />
3x<br />
8<br />
+ =<br />
2 3<br />
x + x = 4<br />
2<br />
2 3 9 9<br />
x + x + = 4 +<br />
2 16 16<br />
2<br />
x<br />
x<br />
⎛ 3 ⎞ 64 9<br />
⎜ x +<br />
4<br />
⎟ = +<br />
⎝ ⎠ 16 16<br />
⎛ 3 ⎞ 73<br />
⎜ x + ⎟ =<br />
⎝ 4 ⎠ 16<br />
3 73<br />
x + = ±<br />
4 16<br />
3 73<br />
x + = ±<br />
4 4<br />
3 73<br />
x = − ±<br />
4 4<br />
− 3 ± 73<br />
x =<br />
4<br />
2<br />
223
<strong>Chapter</strong> 7 Test<br />
SSM: Intermediate Algebra<br />
17. Solve for x when f ( x ) = 0 :<br />
2<br />
3x<br />
8x<br />
1 0<br />
− − ± − −<br />
x =<br />
2(3)<br />
x =<br />
x =<br />
x =<br />
− + =<br />
2<br />
( 8) ( 8) 4(3)(1)<br />
8 ± 52<br />
6<br />
8 ± 2 13<br />
6<br />
4 ± 13<br />
3<br />
⎛<br />
The x-intercepts are<br />
⎜<br />
⎝<br />
⎛ 4 − 13 ⎞<br />
⎜ ,0<br />
3 ⎟<br />
.<br />
⎝ ⎠<br />
18. Solve for x when f(x) = 0<br />
= − − +<br />
2<br />
0 2( x 3) 5<br />
2<br />
2( x 3) 5<br />
2<br />
− =<br />
( x 3)<br />
− =<br />
x − 3 = ±<br />
5<br />
2<br />
5<br />
2<br />
4 + 13 ,0<br />
3<br />
5<br />
x − 3 = ±<br />
2<br />
Rationalize the denomina<strong>to</strong>r:<br />
5 2<br />
x − 3 = ± ⋅<br />
2 2<br />
Solve for x :<br />
x − 3 = ±<br />
10<br />
2<br />
10<br />
x = 3 ±<br />
2<br />
x ≈ 1.42 or x ≈ 4.58<br />
⎞<br />
⎟<br />
and<br />
⎠<br />
So, the x-intercepts are approximately ( 1.42,0 )<br />
and ( 4.58,0 ) . Since these points are symmetric,<br />
the average of the x-coordinates, 1.42 + 4.58 = 3 ,<br />
2<br />
is the x-coordinate of the vertex of f(x). Substitute 3<br />
for x in the function <strong>to</strong> find the y-coordinate of the<br />
vertex:<br />
224<br />
2<br />
h( x ) = −2(3 − 3) + 5 = 5<br />
Therefore, the vertex is (3, 5).<br />
19. Solve for a when the discriminant equals 0:<br />
2<br />
b − 4ac<br />
= 0<br />
2<br />
( 4) 4( a)(4 a) 0<br />
2<br />
16 16a<br />
0<br />
−<br />
a<br />
− − =<br />
2<br />
16a<br />
16<br />
2<br />
− =<br />
= 1<br />
a = ± 1<br />
= −<br />
20. Substitute the given points in<strong>to</strong><br />
2<br />
( 1,4 ): 4 = a( 1) + b( 1)<br />
+ c<br />
( 2,9 ): 9 =<br />
2<br />
( 2) + ( 2)<br />
+<br />
( 3,16 ):16 2<br />
( 3) ( 3)<br />
a b c<br />
= a + b + c<br />
Simplify these equations:<br />
a+ b+ c = 4 1<br />
( )<br />
( )<br />
( )<br />
4a + 2b + c = 9 2<br />
9a + 3b + c = 16 3<br />
2<br />
y = ax + bx + c .<br />
Eliminate c by multiplying both sides of equation<br />
(1) by -1:<br />
−a − b − c = − 4 4<br />
( )<br />
Adding the left sides and right sides of equations<br />
(2) and (4) gives:<br />
3a<br />
+ b = 5 5<br />
( )<br />
Simplify:<br />
8a<br />
+ 2b<br />
= 12 6<br />
( )<br />
Adding the left sides and right sides of equations<br />
(3) and (4) gives:<br />
4a<br />
+ b = 6 7<br />
( )<br />
Eliminate b by multiplying equation (6) by -1 and<br />
add each side <strong>to</strong> the corresponding side of equation<br />
(7):
SSM: Intermediate Algebra<br />
<strong>Chapter</strong> 7 Test<br />
a = 1<br />
Substitute 1 for a in equation (5):<br />
3 1 + b = 5<br />
( )<br />
b = 2<br />
Then, substitute 1 for a and 2 for b in equation (1):<br />
a+ b+ c = 4<br />
1+ 2 + c = 4<br />
c = 1<br />
Therefore, a = 1, b = 2, and c = 1. So, the equation<br />
2<br />
is y = x + 2x<br />
+ 1.<br />
2<br />
21. <strong>Using</strong> the equation y = a( x − h)<br />
+ k , substitute 5<br />
for h and 3 for k since the vertex is (5, 3). Also,<br />
substitute 3 for x and 11 for y since (3,11) lies on<br />
the parabola.<br />
2<br />
y = a( x − h)<br />
+ k<br />
2<br />
11 a(3 5) 3<br />
2<br />
11 a(3 5) 3<br />
2<br />
11 a( 2) 3<br />
11 = 4a<br />
+ 3<br />
8 = 4a<br />
a = 2<br />
= − +<br />
= − +<br />
= − +<br />
c.<br />
2<br />
x − 6x<br />
+ 11 = 3<br />
2<br />
x − 6x<br />
+ 8 = 0<br />
2<br />
( 6) ( 6) 4(1)(8)<br />
− − ± − −<br />
x =<br />
2(1)<br />
6 ± 4<br />
x =<br />
2<br />
6 ± 2<br />
x =<br />
2<br />
8 4<br />
x = or x =<br />
2 2<br />
x = 4 or x = 2<br />
23. a. First, draw a scattergram of the data. The<br />
quadratic regression equation,<br />
2<br />
f ( t) = − 1.14t + 25.26t<br />
− 81.2 , is the best<br />
model for this data.<br />
So, the equation is<br />
y = − +<br />
2<br />
2( x 5) 3<br />
22. a.<br />
b.<br />
2<br />
x − 6x<br />
+ 11 = 1<br />
2<br />
x − 6x<br />
+ 10 = 0<br />
− − ± − −<br />
x =<br />
2(1)<br />
x =<br />
2<br />
( 6) ( 6) 4(1)(10)<br />
6 ± −4<br />
2<br />
Since the square root of a negative is not a real<br />
number, there are no real number solutions.<br />
There is no such value for x.<br />
2<br />
x − 6x<br />
+ 11 = 2<br />
2<br />
x − 6x<br />
+ 9 = 0<br />
− − ± − −<br />
x =<br />
2(1)<br />
x<br />
x<br />
2<br />
( 6) ( 6) 4(1)(9)<br />
6 ± 0<br />
=<br />
2<br />
6<br />
= = 3<br />
2<br />
b.<br />
2<br />
f (15) = − 1.14(15) + 25.26(15) − 81.2 ≈ 41.2<br />
So, 41.2% of television shows will be at least<br />
an hour long in 2005.<br />
c. Solve for t when f(t) = 50<br />
= − + −<br />
2<br />
50 1.14t<br />
25.26t<br />
81.2<br />
= − + −<br />
2<br />
0 1.14t<br />
1.85t<br />
131.2<br />
− ± − − −<br />
t =<br />
2( −1.14)<br />
2<br />
25.26 25.26 4( 1.14)( 131.2)<br />
t ≈ 8.31 or t ≈ 13.8<br />
So, according <strong>to</strong> the model, 50% of television<br />
shows will be at least an hour long in 1996 and<br />
in 2004.<br />
225
Cumulative Review <strong>Chapter</strong>s 1-7<br />
SSM: Intermediate Algebra<br />
d. Solve for t when f(t) = 0<br />
= − + −<br />
2<br />
0 1.14t<br />
25.26t<br />
81.2<br />
2<br />
25.26 25.26 4( 1.14)( 81.2)<br />
− ± − − −<br />
t =<br />
2( −1.14)<br />
t ≈ 3.90 or t ≈ 18.26<br />
So, the t-intercepts are ( 3.90,0 ) and<br />
( 18.26,0 ). This means that according <strong>to</strong> the<br />
model no television shows were at least an<br />
hour in 1994 and none will be at least an hour<br />
in 2008. (<strong>Model</strong> breakdown has probably<br />
occurred.<br />
e. <strong>Model</strong> breakdown will definitely occur for<br />
years before 1994 (t < 3.90) and for years after<br />
2008 (t > 18.26) because there can’t be less<br />
than zero shows that last at least an hour.<br />
24. <strong>Using</strong> a graphing calcula<strong>to</strong>r, find that the<br />
maximum point of the parabola is approximately<br />
(2.50, 103).<br />
So, the maximum height reached by the ball is 103<br />
feet at 2.5 seconds.<br />
Cumulative Review of <strong>Chapter</strong>s 1-7<br />
1. 5x 2 = 24<br />
2 24<br />
x =<br />
5<br />
= ±<br />
= ±<br />
24<br />
5<br />
24<br />
5<br />
2 6 5<br />
= ± ⋅<br />
5 5<br />
= ±<br />
2 30<br />
5<br />
2. 5 x + 1 =<br />
7<br />
6 3 2<br />
5 7 1<br />
x = −<br />
6 2 3<br />
5 21 2<br />
x = −<br />
6 6 6<br />
5 19<br />
x =<br />
6 6<br />
6 5 19 6<br />
⋅ x = ⋅<br />
5 6 6 5<br />
19<br />
x =<br />
5<br />
3. log<br />
4<br />
( x + 3) = 2<br />
x + 3 = 4<br />
2<br />
x + 3 = 16<br />
x = 13<br />
x<br />
4. 3e − 5 = 49<br />
x<br />
3e<br />
= 54<br />
x 54<br />
e =<br />
3<br />
x ⎛ 54 ⎞<br />
ln( e ) = ln ⎜ ⎟<br />
⎝ 3 ⎠<br />
⎛ 54 ⎞<br />
x ln( e) = ln ⎜ ⎟<br />
⎝ 3 ⎠<br />
⎛ 54 ⎞<br />
x = ln ⎜ ⎟ ≈ 2.8904<br />
⎝ 3 ⎠<br />
5. 5b 6 + 4 = 82<br />
6<br />
5b<br />
78<br />
b<br />
6<br />
=<br />
=<br />
78<br />
5<br />
1<br />
6<br />
⎛ 78 ⎞<br />
b = ± ⎜ ⎟ ≈ ± 1.5807<br />
⎝ 5 ⎠<br />
6. 3x<br />
2 − 5x<br />
− 4 = 0<br />
− − ± − − −<br />
x =<br />
2(3)<br />
2<br />
( 5) ( 5) 4(3)( 4)<br />
5 ± 73<br />
x =<br />
6<br />
226
SSM: Intermediate Algebra Cumulative Review <strong>Chapter</strong>s 1-7<br />
7.<br />
8.<br />
4x−5<br />
3(2) 95<br />
(2)<br />
4 x−5<br />
=<br />
95<br />
=<br />
3<br />
⎛ 95 ⎞<br />
= ⎜ ⎟<br />
4x−5<br />
log(2) log<br />
⎝ 3 ⎠<br />
95<br />
⎛ ⎞<br />
(4x<br />
− 5)log(2) = log⎜ ⎟<br />
⎝ 3 ⎠<br />
⎛ 95 ⎞<br />
log ⎜ ⎟ 3<br />
4x<br />
− 5 =<br />
⎝ ⎠<br />
log(2)<br />
⎛ 95 ⎞<br />
log ⎜ ⎟ 3<br />
4x<br />
=<br />
⎝ ⎠<br />
+ 5<br />
log(2)<br />
⎛ 95 ⎞<br />
log ⎜ ⎟<br />
⎝ 3 ⎠ + 5<br />
log(2)<br />
x = ≈ 2.4962<br />
4<br />
2<br />
2(3x<br />
− 10) = − 7x<br />
2<br />
6 20 7<br />
x<br />
2<br />
6 7 20 0<br />
x<br />
−<br />
= −<br />
(3x<br />
− 4)(2x<br />
+ 5) = 0<br />
3x<br />
− 4 = 0 or 2x<br />
+ 5 = 0<br />
x<br />
+ x − =<br />
3x<br />
= 4 or 2x<br />
= −5<br />
4 5<br />
x = or x = −<br />
3 2<br />
9. log<br />
b<br />
(65) = 4<br />
b<br />
4<br />
= 65<br />
1<br />
4<br />
b = 65<br />
b ≈ 2.8394<br />
10. 3x 2 + 5x 2 = 12x<br />
+ 20<br />
2<br />
8 12 20 0<br />
x<br />
− x − =<br />
2<br />
4(2 3 5) 0<br />
x<br />
− x − =<br />
2<br />
2 3 5 0<br />
x<br />
− x − =<br />
(2x<br />
− 5)( x + 1) = 0<br />
2x<br />
− 5 = 0 or x + 1 = 0<br />
2x<br />
= 5 or x = −1<br />
5<br />
x = or x = − 1<br />
2<br />
11. 2(5x<br />
+ 2) − 1 = 9( x − 3)<br />
10x<br />
+ 4 − 1 = 9x<br />
− 27<br />
12.<br />
13.<br />
14.<br />
10x<br />
+ 3 = 9x<br />
− 27<br />
x = −30<br />
3log (4 x ) − 2log (4 x ) = 5<br />
4 3<br />
2 2<br />
log (4 x ) − log (4 x ) = 5<br />
4 3 3 2<br />
2 2<br />
log (64 x ) − log (16 x ) = 5<br />
12 6<br />
2 2<br />
12<br />
⎛ 64x<br />
⎞<br />
2 ⎜ 6 ⎟ =<br />
log 5<br />
⎝ 16x<br />
⎠<br />
2<br />
6<br />
( x )<br />
log 4 = 5<br />
4x<br />
= 2<br />
6 5<br />
6<br />
4 32<br />
x<br />
x<br />
6<br />
=<br />
= 8<br />
x = ± (8)<br />
1<br />
6<br />
x ≈ 1.4142<br />
2 x(3x − 4) + 5 = 3−<br />
x<br />
6x − 8x + 5 = 3 − x<br />
2 2<br />
2<br />
7 8 2 0<br />
x<br />
− x + =<br />
− − ± − −<br />
x =<br />
2(7)<br />
x =<br />
x =<br />
x =<br />
2<br />
( 8) ( 8) 4(7)(2)<br />
8 ± 8<br />
14<br />
8 ± 2 2<br />
14<br />
4 ± 2<br />
7<br />
4 7<br />
3ln(2 ) 2ln(3 ) 8<br />
x<br />
2<br />
+ x =<br />
4 3 7 2<br />
ln(2 ) ln(3 ) 8<br />
12 14<br />
ln(8 ) ln(9 ) 8<br />
26<br />
ln(72 ) 8<br />
72x<br />
x<br />
26<br />
x<br />
x<br />
x<br />
26 8<br />
8<br />
e<br />
=<br />
72<br />
+ x =<br />
+ x =<br />
= e<br />
=<br />
1<br />
8 26<br />
⎛ e ⎞<br />
x = ⎜ ⎟ ≈ 1.1540<br />
⎝ 72 ⎠<br />
227
Cumulative Review <strong>Chapter</strong>s 1-7<br />
SSM: Intermediate Algebra<br />
15.<br />
2<br />
5( x − 3) + 4 = 13<br />
2<br />
5( x 3) 9<br />
2<br />
− =<br />
( x 3)<br />
9<br />
5<br />
x − 3 = ±<br />
9<br />
5<br />
x − 3 = ±<br />
3<br />
5<br />
Rationalize the denomina<strong>to</strong>r:<br />
3 5<br />
x − 3 = ± ⋅<br />
5 5<br />
3 5<br />
x − 3 = ±<br />
5<br />
3 5<br />
x = 3 ±<br />
5<br />
x =<br />
− =<br />
15 ± 3 5<br />
5<br />
37x<br />
= 37<br />
x = 1<br />
Substitute 1 for x in one of the original equations <strong>to</strong><br />
solve for y:<br />
4(1) − 7 y = −10<br />
− 7 y = −14<br />
y = 2<br />
The solution is (1, 2).<br />
18. Substitute y = 3x<br />
− 1 in 2x<br />
− 3y<br />
= − 11<br />
2x<br />
− 3(3x<br />
− 1) = −11<br />
2x<br />
− 9x<br />
+ 3 = −11<br />
− 7x<br />
= −14<br />
x = 2<br />
Substitute 2 for x in one of the original equations <strong>to</strong><br />
solve for y:<br />
y = 3x<br />
−1<br />
= 3(2) −1<br />
= 5<br />
16. 2x<br />
2 + 3x<br />
− 6 = 0<br />
2<br />
2x<br />
+ 3x<br />
= 6<br />
2 3<br />
x + x = 3<br />
2<br />
2 3 9 9<br />
x + x + = 3 +<br />
2 16 16<br />
2<br />
⎛ 3 ⎞ 48 9<br />
⎜ x + ⎟ = +<br />
⎝ 4 ⎠ 16 16<br />
⎛ 3 ⎞ 57<br />
⎜ x + ⎟ =<br />
⎝ 4 ⎠ 16<br />
3 57<br />
x + = ±<br />
4 16<br />
3 57<br />
x + = ±<br />
4 4<br />
3 57<br />
x = − ±<br />
4 4<br />
x =<br />
2<br />
− 3 ± 57<br />
4<br />
17. In order <strong>to</strong> eliminate y, multiply 4x<br />
− 7 y = − 10 by<br />
4 and 3x<br />
+ 4y<br />
= 11 by 7. This gives:<br />
16x<br />
− 28y<br />
= −40<br />
21x<br />
+ 28y<br />
= 77<br />
Add corresponding sides <strong>to</strong> obtain x:<br />
The solution is (2, 5).<br />
19. In order <strong>to</strong> eliminate y, multiply 1 x − y = 5 by<br />
2 2<br />
3<br />
3 3 15<br />
− . This gives − x + y = − . Add the left<br />
5<br />
10 5 10<br />
sides and the right sides of the equations:<br />
3 3 15<br />
− x + y = −<br />
10 5 10<br />
2 3 6<br />
x − y =<br />
5 5 5<br />
This yields:<br />
3 2 15 6<br />
− x + x = − +<br />
10 5 10 5<br />
3 4 15 12<br />
− x + x = − +<br />
10 10 10 10<br />
1 3<br />
x = −<br />
10 10<br />
10 1 3 10<br />
⋅ x = − ⋅<br />
1 10 10 1<br />
x = −3<br />
Substitute -3 for x in one of the original equations<br />
<strong>to</strong> solve for y:<br />
228
SSM: Intermediate Algebra Cumulative Review <strong>Chapter</strong>s 1-7<br />
1 5<br />
( −3)<br />
− y =<br />
2 2<br />
3 5<br />
− − y =<br />
2 2<br />
3 5<br />
y = − −<br />
2 2<br />
8<br />
y = − = −4<br />
2<br />
So the solution is (-3, -4).<br />
20. 2(3x<br />
− 4) < 5 − 3(6x<br />
+ 5)<br />
2(3x<br />
− 4) < 5 − 3(6x<br />
+ 5)<br />
21.<br />
22.<br />
6x<br />
− 8 < 5 −18x<br />
−15<br />
6x<br />
− 8 < −18x<br />
−10<br />
24x<br />
< −2<br />
1<br />
x < −<br />
12<br />
⎛ 1 ⎞<br />
⎜ −∞,<br />
− ⎟<br />
⎝ 12 ⎠<br />
(2 b c ) (3 b c )<br />
=<br />
4 −5 3 −1 −2 4<br />
3 12 −15 4 −4 −8<br />
(2 b c )(3 b c )<br />
= ⋅<br />
12 −4 −15 −8<br />
(8 81)( b b )( c c )<br />
= 648b c<br />
648b<br />
=<br />
23<br />
c<br />
10b<br />
15b<br />
2b<br />
=<br />
2b<br />
=<br />
2b<br />
=<br />
=<br />
8 −23<br />
8<br />
3 2<br />
−<br />
4 5<br />
c<br />
1 3<br />
−<br />
2 5<br />
c<br />
3 ⎛ 1 ⎞<br />
− −<br />
2 3<br />
⎜ − ⎟ −<br />
4 ⎝ 2 ⎠ 5 5<br />
3 2 1<br />
− + −<br />
4 4 5<br />
1 1<br />
− −<br />
4 5<br />
3<br />
2<br />
1 1<br />
4 5<br />
3b c<br />
3<br />
c<br />
3<br />
c<br />
c<br />
1<br />
−<br />
12<br />
x<br />
23.<br />
24.<br />
25.<br />
26.<br />
⎛ 6b c<br />
⎜<br />
⎝ 8b c<br />
8 −3<br />
−4 −1<br />
⎛ 6b c<br />
⎜<br />
⎝ 8b c<br />
8 −3<br />
−4 −1<br />
36<br />
6<br />
3<br />
⎞<br />
⎟<br />
⎠<br />
8 −( −4) −3 −( −1)<br />
⎛ 3b<br />
c ⎞<br />
= ⎜ ⎟<br />
⎝ 4 ⎠<br />
3<br />
3<br />
12 −2<br />
⎛ 3b c ⎞<br />
= ⎜ ⎟<br />
⎝ 4 ⎠<br />
12<br />
⎛ 3b<br />
⎞<br />
= ⎜ 2 ⎟<br />
⎝ 4c<br />
⎠<br />
27b<br />
=<br />
64c<br />
⎞<br />
⎟<br />
⎠<br />
+<br />
7 5<br />
3ln(2 x ) 2ln(4 x )<br />
= ln(2 x ) + ln(4 x )<br />
= +<br />
3<br />
3<br />
7 3 5 2<br />
21 10<br />
ln(8 x ) ln(16 x )<br />
= ⋅<br />
=<br />
21 10<br />
ln(8x<br />
16 x )<br />
31<br />
ln(128 x )<br />
−<br />
8 3<br />
3log<br />
b<br />
(4 x ) 4log<br />
b<br />
(2 x )<br />
= log (4 x ) − log (2 x )<br />
b<br />
= −<br />
8 3 3 4<br />
b<br />
24 12<br />
log<br />
b<br />
(64 x ) log<br />
b<br />
(16 x )<br />
24<br />
⎛ 64x<br />
⎞<br />
= logb<br />
⎜ 12 ⎟<br />
⎝ 16x<br />
⎠<br />
= log 4<br />
b<br />
12<br />
( x )<br />
2<br />
(3x 4) (3x 4)(3x<br />
4)<br />
− = − −<br />
= − − +<br />
2<br />
9x 12x 12x<br />
16<br />
= − +<br />
2<br />
9x<br />
24x<br />
16<br />
27. (5x<br />
− 7)(5x<br />
+ 7)<br />
= + − −<br />
2<br />
25x 35x 35x<br />
49<br />
2<br />
= 25x<br />
− 49<br />
2 3<br />
28. −3 x( x − 5)( x + 8)<br />
= − + − −<br />
5 2 3<br />
3 x( x 8x 5x<br />
40)<br />
= − − + −<br />
5 3 2<br />
3 x( x 5x 8x<br />
40)<br />
6 4 3<br />
= − 3x + 15x − 24x + 120x<br />
229
Cumulative Review <strong>Chapter</strong>s 1-7<br />
SSM: Intermediate Algebra<br />
29.<br />
30.<br />
31.<br />
2<br />
(2x 3)( x 4x<br />
5)<br />
− + −<br />
= + − − − +<br />
3 2 2<br />
2x 8x 10x 3x 12x<br />
15<br />
= + − +<br />
3 2<br />
2x 5x 22x<br />
15<br />
f x<br />
= − − +<br />
2<br />
( ) 2( x 5) 3<br />
= −2( x − 5)( x − 5) + 3<br />
= − − − + +<br />
2<br />
2( x 5x 5x<br />
25) 3<br />
= − − + +<br />
2<br />
2( x 10x<br />
25) 3<br />
= − + − +<br />
2<br />
2x<br />
20x<br />
50 3<br />
= − + −<br />
2<br />
2x<br />
20x<br />
47<br />
− = − = − +<br />
2 2 2<br />
81x 25 (9 x) 5 (9x 5)(9x<br />
5)<br />
40. h − 1 (7) = 4 since h (4) = 7<br />
41. The x-intercept of k is (2, 0)<br />
42.<br />
32. x 3 − 13x 2 + 40x<br />
2<br />
= x( x − 13x<br />
+ 40)<br />
= x( x − 8)( x − 5)<br />
43.<br />
2<br />
33. 8x + 22x − 21 = (2x + 7)(4x<br />
− 3)<br />
34. x 3 + 4x 2 − 9x− 36 :<br />
2<br />
= x ( x + 4) − 9( x + 4)<br />
= + −<br />
2<br />
( x 4)( x 9)<br />
= ( x + 4)( x − 3)( x + 3)<br />
35. Since the y-intercept is (0, 20) and as x increases<br />
by 1, f(x) decreases by 3: f ( x) = − 3x<br />
+ 20<br />
36. As x increases by 1, g(x) increases by a fac<strong>to</strong>r of 3,<br />
so the base b = 3. Substitute a point from the table<br />
x<br />
in<strong>to</strong> g( x)<br />
= ab and solve for a.<br />
12 = a(3)<br />
12 = 9a<br />
2<br />
4<br />
a =<br />
3<br />
So, the equation is:<br />
4<br />
g( x) = (3)<br />
3<br />
x<br />
44. First, change the form in<strong>to</strong> y = mx + b<br />
2x<br />
− 5y<br />
= 20<br />
− 5y<br />
= − 2x<br />
+ 20<br />
2<br />
y = x − 4<br />
5<br />
37. For the function k, as x increases by 1, k(x)<br />
increases by 4, so the slope is 4.<br />
38. g (4) = 108<br />
39. x = 1 since g (1) = 4<br />
230
SSM: Intermediate Algebra Cumulative Review <strong>Chapter</strong>s 1-7<br />
45.<br />
49.<br />
y − y<br />
m x − x<br />
2 1<br />
= = = =<br />
2 1<br />
4 − ( −2) 6<br />
−3 − ( −5) 2<br />
3<br />
46.<br />
50. First, change the form of the equation given <strong>to</strong><br />
y = mx + b .<br />
3x<br />
− 4y<br />
= 5<br />
− 4y<br />
= − 3x<br />
+ 5<br />
3 5<br />
y = x −<br />
4 4<br />
4<br />
The slope of the perpendicular line will be − . So<br />
3<br />
the equation of the perpendicular line will be<br />
4<br />
y = − x + b . Substitute the given point in<strong>to</strong> this<br />
3<br />
equation <strong>to</strong> solve for b.<br />
4<br />
6 = − ( − 2) + b<br />
3<br />
8<br />
6 = + b<br />
3<br />
8<br />
b = 6 − 3<br />
47.<br />
48.<br />
18 8<br />
b = −<br />
3 3<br />
10<br />
b =<br />
3<br />
4 10<br />
So, the equation is y = − x + or<br />
3 3<br />
y = − 1.33x<br />
+ 3.33 .<br />
x<br />
51. The equation is of the form y = ab . Since (0, 78)<br />
is on the curve, a = 78. So, the equation is of the<br />
form: y = 78b<br />
x . Now substitute (9, 13) in<strong>to</strong> the<br />
equation and solve for b.<br />
9<br />
13 = 78b<br />
b<br />
9<br />
=<br />
13<br />
78<br />
1<br />
9<br />
⎛ 13 ⎞<br />
b = ⎜ ⎟ ≈ 0.82<br />
⎝ 78 ⎠<br />
So, the equation is<br />
y = 78(0.82) x .<br />
52. Both points satisfy the equation<br />
the system of equations:<br />
5<br />
83 = ab<br />
27 = ab<br />
2<br />
y<br />
x<br />
= ab ; we have<br />
231
Cumulative Review <strong>Chapter</strong>s 1-7<br />
SSM: Intermediate Algebra<br />
Combining these two equations yields<br />
5<br />
83 ab<br />
=<br />
2<br />
27 ab<br />
83 3<br />
= b<br />
27<br />
1<br />
3<br />
⎛ 83 ⎞<br />
b = ⎜ ⎟ ≈ 1.45<br />
⎝ 27 ⎠<br />
So the equation is of the form:<br />
To find a, substitute (2, 27) in<strong>to</strong><br />
27 = a(1.45)<br />
2<br />
27 = 2.1025a<br />
a = 12.84<br />
So, the equation is<br />
y = 12.84(1.45) x<br />
53. Substitute the given points in<strong>to</strong><br />
( 1, −1 ): − 1 =<br />
2<br />
( 1) + ( 1)<br />
+<br />
( 2, 4 ): 4 =<br />
2<br />
( 2) + ( 2)<br />
+<br />
( 4, 20 ): 20<br />
2<br />
( 4) ( 4)<br />
a b c<br />
a b c<br />
= a + b + c<br />
Simplify these equations:<br />
a+ b+ c = −1 1<br />
( )<br />
( )<br />
( )<br />
4a + 2b + c = 4 2<br />
16a + 4b + c = 20 3<br />
y = a(1.45) x<br />
y = a(1.45) x<br />
2<br />
y = ax + bx + c .<br />
Eliminate c by multiplying both sides of equation<br />
(1) by -1:<br />
−a − b − c = 1 4<br />
( )<br />
Adding the left sides and right sides of equations<br />
(2) and (4) gives:<br />
3a<br />
+ b = 5 5<br />
( )<br />
Adding the left sides and right sides of equations<br />
(3) and (4) gives:<br />
15a<br />
+ 3b<br />
= 21 6<br />
Simplify:<br />
5a<br />
+ b = 7 7<br />
( )<br />
( )<br />
Eliminate b by multiplying equation (5) by -1 and<br />
add each side <strong>to</strong> the corresponding side of equation<br />
(7):<br />
2a<br />
= 2<br />
a = 1<br />
Next, substitute 1 for a in equation (5):<br />
3 1 + b = 5<br />
( )<br />
b = 2<br />
Then, substitute 1 for a and 2 for b in equation (1):<br />
a+ b+ c = −1<br />
1+ 2 + c = −1<br />
c = −4<br />
Therefore, a = 1, b = 2, and c = -4. So, the equation<br />
2<br />
is y = x + 2x<br />
− 4 .<br />
54. a. Linear:<br />
b.<br />
6 − 3<br />
slope = m = = 3<br />
1−<br />
0<br />
y − intercept = (0,3)<br />
So, y = 3x<br />
+ 3<br />
Exponential:<br />
As the value of x increases by 1, the value of y<br />
is multiplied by 2,, so the base b = 2. The y-<br />
intercept is (0, 3). Therefore, the exponential<br />
function is y = 3( 2)<br />
x<br />
<strong>Quadratic</strong>:<br />
Answers may vary. Example:<br />
55. − x<br />
2 + 6x<br />
− 5 = 3<br />
2<br />
x − 6x<br />
+ 8 = 0<br />
( x − 4)( x − 2) = 0<br />
x − 4 = 0 or x − 2 = 0<br />
x = 4 or x = 2<br />
56. − x<br />
2 + 6x<br />
− 5 = 4<br />
2<br />
x − 6x<br />
+ 9 = 0<br />
( x − 3)( x − 3) = 0<br />
− =<br />
2<br />
( x 3) 0<br />
x − 3 = 0<br />
x = 3<br />
y<br />
2<br />
= 3x<br />
+ 3 .<br />
232
SSM: Intermediate Algebra Cumulative Review <strong>Chapter</strong>s 1-7<br />
57. − x<br />
2 + 6x<br />
− 5 = 5<br />
65.<br />
2<br />
x − 6x<br />
+ 10 = 0<br />
x =<br />
± − −<br />
2<br />
6 ( 6) 4(1)(10)<br />
2(1)<br />
6 ± −4<br />
x =<br />
2<br />
Since the square root of a negative is not a real<br />
number, there are no real number solutions. There<br />
is no such value for x.<br />
58. To find the x-intercepts, let f(x) = 0 and solve for x:<br />
2<br />
− + − =<br />
x<br />
6x<br />
5 0<br />
2<br />
x − 6x<br />
+ 5 = 0<br />
( x − 5)( x − 1) = 0<br />
x − 5 = 0 or x − 1 = 0<br />
x = 5 or x = 1<br />
The x-intercepts are (5, 0) and (1, 0).<br />
59. To find the y-intercept, find f(0)<br />
60.<br />
2<br />
f (0) = − (0) + 6(0) − 5 = − 5<br />
The y-intercept is (0. -5)<br />
61. log<br />
2<br />
(16) = 4 since<br />
62.<br />
4<br />
2 = 16<br />
3<br />
log ( b b<br />
) = 3 since 3 3<br />
b = b<br />
66. g( x ) = 3 x<br />
67.<br />
−1<br />
g x =<br />
3<br />
x<br />
( ) log ( )<br />
2<br />
f ( x) = x + 1<br />
5<br />
Replace f ( x) with y :<br />
2<br />
y = x + 1<br />
5<br />
Solve for x :<br />
2<br />
y − 1 = x<br />
5<br />
5 ( y − 1) = x<br />
2<br />
5 5<br />
x = y −<br />
2 2<br />
−1<br />
Replace x with g ( y) :<br />
−1<br />
5 5<br />
g ( y)<br />
= y −<br />
2 2<br />
Write in terms of x :<br />
−1<br />
5 5<br />
g ( x)<br />
= x −<br />
2 2<br />
68. a. First, draw a scattergram of the data. The<br />
linear regression equation,<br />
f ( t) = 3.97t<br />
+ 3.02 , is the best model for this<br />
data.<br />
1 1<br />
log ⎜ ⎟ = log ⎜ = log 5 = −2<br />
2 ⎟<br />
⎝ 25 ⎠ ⎝ 5 ⎠<br />
⎛ ⎞ ⎛ ⎞<br />
−2<br />
63.<br />
5 5 5 ( )<br />
log(2)<br />
64. log<br />
9<br />
(2) = ≈ 0.3155<br />
log(9)<br />
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Cumulative Review <strong>Chapter</strong>s 1-7<br />
SSM: Intermediate Algebra<br />
b. f (18) = 3.97(18) + 3.02 = 74.48 . In 2008,<br />
74.48% of companies will offer s<strong>to</strong>ck options<br />
<strong>to</strong> at least half of their employees.<br />
c. 100 = 3.97t<br />
+ 3.02<br />
96.98 = 3.97t<br />
t ≈ 24.43<br />
All companies will offer s<strong>to</strong>ck options <strong>to</strong> at<br />
least half of their employees in 2014.<br />
d. Set f(t) = 0 and solve for t<br />
0 = 3.97t<br />
+ 3.02<br />
− 3.02 = 3.97t<br />
3.02<br />
t = − ≈ −0.76<br />
3.97<br />
So, the t-intercept is (-0.76, 0). This means<br />
that no companies offered s<strong>to</strong>ck options <strong>to</strong> at<br />
least half of their employees in 1989. <strong>Model</strong><br />
breakdown has occurred.<br />
e. t < -0.76 (since there can’t be less than 0<br />
percent offering s<strong>to</strong>ck options) or t> 24.43<br />
(because there can’t be more than 100 percent<br />
of companies offering s<strong>to</strong>ck options).<br />
69. a. Find the different models using the regression<br />
feature on a graphing calcula<strong>to</strong>r:<br />
Linear: f ( t) = 0.18t<br />
− 0.61<br />
Exponential: f ( t ) = 0.15(1.22) t<br />
<strong>Quadratic</strong>:<br />
f t = t − t +<br />
2<br />
( ) 0.0216 0.19 0.88<br />
b. Answers may vary. Example: both the<br />
quadratic and exponential functions model the<br />
data well.<br />
c. <strong>Quadratic</strong> model<br />
d. <strong>Using</strong> a graphing calcula<strong>to</strong>r, find that the<br />
minimum point of the parabola is<br />
approximately (4.41, 0.46).<br />
e. Since the base of the exponential model is b =<br />
1.22, the rate of growth is approximately 1.22-<br />
1 = 0.22 or 22% per year.<br />
f. <strong>Quadratic</strong> model:<br />
2<br />
5 0.0216t<br />
0.19t<br />
0.88<br />
2<br />
0 0.0216t<br />
0.19t<br />
4.12<br />
t =<br />
= − +<br />
= − −<br />
± − − −<br />
2<br />
0.19 ( 0.19) 4(0.0216)( 4.12)<br />
t ≈ −10.1 or t ≈ 18.9<br />
2(0.0216)<br />
t = -10.1 is model breakdown, so the year that<br />
the quadratic model predicts that sakes will<br />
reach $5 billion in 2009.<br />
Exponential model:<br />
t<br />
5 = 0.15(1.22)<br />
t<br />
33.3333 = 1.22<br />
t<br />
log(1.22) = log(33.33333)<br />
t log(1.22) = log(33.33333)<br />
t =<br />
log(33.33333)<br />
log(1.22)<br />
t ≈ 17.6<br />
So the year that the exponential model<br />
predicts that sakes will reach $5 billion in<br />
2008.<br />
Answers may vary. Example:<br />
It makes sense that the year predicted by the<br />
exponential model is before the year predicted<br />
by the quadratic model, as the exponential<br />
model is a steeper curve, and rises quicker<br />
than the quadratic model.<br />
70. a. First, draw a scattergram of the data for h. The<br />
linear regression equation, h( t) = 2.4t<br />
+ 10.7 ,<br />
is the best model for this data.<br />
This means that the minimum sales revenues<br />
was 0.46 billion dollars.<br />
234
SSM: Intermediate Algebra Cumulative Review <strong>Chapter</strong>s 1-7<br />
First, draw a scattergram of the data for c. The<br />
linear regression equation, c( t) = t + 53 , is the<br />
best model for this data.<br />
b. The rates of change are the slopes of the<br />
respective models.<br />
Rate of change at Hartford: 2.4 percent per<br />
year.<br />
Rate of change in Connecticut: 1 percent per<br />
year.<br />
c. h (18) = 2.4(18) + 10.7 = 53.9 . This means that<br />
in 2008, the percent of students who will score<br />
above the goals in Hartford will be 53.9%.<br />
c (18) = 18 + 53 = 71. This means that in 2008,<br />
the percent of students who will score above<br />
the goals in Connecticut will be 71%.<br />
d. Solve for t when h( t) = c( t)<br />
2.4t<br />
+ 10.7 = t + 53<br />
1.4t<br />
= 42.3<br />
t ≈ 30.21<br />
So, the percentage of Hartford and<br />
Connecticut students who will score above the<br />
goals will be equal in 2020.<br />
235