College Algebra & Trigonometry, 2018a

1.3. QUADRATIC EQUATIONS 31
If we re-examine the sample perfect binomial squares from the previous page, we
note a useful pattern. This is that the blank in the parentheses (x+ ) 2 is filled
by a number that is one-half the value of the linear coefficient - or the coefficient
of the x 1 term. Notice that in x 2 +6x +9=(x +3) 2 , 3 is half of 6, in x 2 +8x +16=
(x +4) 2 , the 4 is half of 8, and so on. If we want to write x 2 + b x+ as a perfect
a
square in the form (x+ ) 2 , the blank in the parentheses should be filled by:
1
2 ∗ b a = b
2a
Now, it’s not true that x 2 + b (
a x+ = x + b ) 2
2a
We’re missing the constant term on the left. However, if we return to our perfect
square examples, we can see that the constant term is always the square of the
term inside inside the parentheses. So, we can restate our problem now as:
x 2 + b ( ) 2 ( b
a x + = x + b ) 2
2a
2a
x 2 + b (
a x + b2
4a = x + b ) 2
2 2a
So, if we return to our originial problem, we were saying that:
x 2 + b a x + c a =0
− c a = − c a
x 2 + b a x
= − c a