Question

some of the steps in the derivation of the quadratic formula are shown. step 3: $-c + \frac{b^2}{4a} = a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}\
ight)$ step 4a: $-c + \frac{b^2}{4a} = a\left(x + \frac{b}{2a}\
ight)^2$ step 4b: $-\frac{4ac}{4a} + \frac{b^2}{4a} = a\left(x + \frac{b}{2a}\
ight)^2$ which best explains or justifies step 4b? \bigcirc factoring a polynomial \bigcirc multiplication property of equality \bigcirc converting to a common denominator \bigcirc addition property of equality

Explanation:

Step1: Analyze Step 4b Change

In Step 4a, we have \(-c + \frac{b^2}{4a}\), and in Step 4b, it becomes \(-\frac{4ac}{4a}+\frac{b^2}{4a}\). We need to see how \(-c\) became \(-\frac{4ac}{4a}\).

Step2: Check Common Denominator

To convert \(-c\) to a fraction with denominator \(4a\), we multiply numerator and denominator by \(4a\) (since \(c=\frac{4ac}{4a}\) when we get a common denominator). So we are converting \(-c\) to have the same denominator as \(\frac{b^2}{4a}\), which is converting to a common denominator.

Answer:

converting to a common denominator