Algebra Know-It-ALL

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.

6. 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

7. 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