This equation is equivalent to the other one. To show this, we can derive one squared
binomial from the other:
(4x− 5)^2 = (4x− 5)(4x− 5)
= (−1)^2 (4x− 5)(4x− 5)
= (−1)(4x− 5)(−1)(4x− 5)
= (− 4 x+ 5)(− 4 x+ 5)
= (− 4 x+ 5)^2
This duplicity occurs with all squared binomials. It simply comes out of the fact that
(−1)^2 = 1.
- To find the root of the quadratic, we can start with either of the binomial factor
equations we found. Let’s use the first one:
(4x− 5)^2 = 0
Taking the square root of both sides, we obtain
4 x− 5 = 0
We can add 5 to each side, getting
4 x= 5
Dividing through by 4 gives us the root x= 5/4. The solution set is {5/4}. Let’s plug the
root into the original quadratic to be sure that it works:
16 x^2 − 40 x+ 25 = 0
16 × (5/4)^2 − 40 × (5/4) + 25 = 0
16 × 25/16 − 50 + 25 = 0
25 − 50 + 25 = 0
− 25 + 25 = 0
0 = 0
- We want to morph the following quadratic so we can get the left side into a product of
a binomial with itself.
x^2 + 6 x− 7 = 0
We can add 16 to each side to get
x^2 + 6 x+ 9 = 16
The left side can now be factored into a square of a binomial:
(x+ 3)^2 = 16
Chapter 22 673