Chapter 9 Radical Functions

Homework 9.3
SSM: Intermediate Algebra
49.
2 x 2 x 3 x + 7
= ⋅
3 x − 7 3 x − 7 3 x + 7
2 x ⋅ 3 x + 2 x ⋅7
=
( 3 x ) − ( 7)
2
( x ) + x
2
2
( x ) −
2 2
6 14
=
3 49
57.
6 x + 5 6 x + 5 3 x + 7
= ⋅
3 x − 7 3 x − 7 3 x + 7
6 x ⋅ 3 x + 6 x ⋅ 7 + 5 ⋅ 3 x + 5 ⋅ 7
=
( )
2
( 3 x ) − ( 7)
( x )
2 2
18 x + 6 7 ⋅ x + 3 5⋅ x + 5⋅7
=
2
2
3 − 7
6x
+ 14 x
=
9x
− 49
18x + 6 7x + 3 5x
+ 35
=
9x
− 7
51.
53.
x − 5 x − 5 x − 5
= ⋅
x + 5 x + 5 x − 5
=
( x ) − 2( x )( 5) + ( 5)
2 2
( x ) − ( 5)
2 2
x − 10 x + 25
=
x − 25
x + 3 x + 3 4 + x
= ⋅
4 − x 4 − x 4 + x
=
x ⋅ 4 + x ⋅ x + 3⋅ 4 + 3⋅
x
2
( 4) − ( x ) 2
4 x + x + 12 + 3 x
=
16 − x
x + 7 x + 12
=
16 − x
59. Student 1 did the work correctly. Student 2’s
error was to square the entire expression. This
changes the value of the expression.
61. Answers may vary. The student did not
rationalize the denominator correctly. For the
radicand in the denominator to be a perfect cube,
2
x needs to be multiplied by x to yield
3 2 3 3
3 x ⋅ x = x .
3 2
5 5 x
= ⋅
x x x
3 3 3 2
=
5
5
=
3 2
x
3 2
x ⋅ x
3 2
5 x
=
3 3
x
3 2
x
x
55.
2 x + 5 2 x + 5 3 x + 1
= ⋅
3 x −1 3 x − 1 3 x + 1
2 x ⋅ 3 x + 2 x ⋅ 1+ 5⋅ 3 x + 5⋅1
=
( )
2
( 3 x ) −( 1)
( x )
2 2
6 x + 2 x + 15 x + 5
=
2
2
3 −1
63.
x x x
= ⋅
3 3 x
2
x
=
3 x
x
=
3 x
6x
+ 17 x + 5
=
9x
−1
284