Step A:
[tex]\[a\left(x+\frac{b}{2a}\right)^2=-c+\frac{b^2}{4a}\][/tex]
[tex]\[a\left(x+\frac{b}{2a}\right)^2=\frac{-4ac+b^2}{4a}\][/tex]

Justification of step A: common denominator

Step B:
[tex]\[a\left(x+\frac{b}{2a}\right)^2=\frac{-4ac+b^2}{4a}\][/tex]
[tex]\[\left(\frac{1}{a}\right)a\left(x+\frac{b}{2a}\right)^2=\left(\frac{1}{a}\right)\left(\frac{b^2-4ac}{4a}\right)\][/tex]

Justification of step B:
- distributive property
- addition property of equality
- multiplication property of equality



Answer :

Sure, let's go through the steps and provide the justifications.

For Step A:
[tex]\[ a\left(x+\frac{b}{2a}\right)^2 = -c + \frac{b^2}{4a} \][/tex]
[tex]\[ a\left(x+\frac{b}{2a}\right)^2 = \frac{-4ac + b^2}{4a} \][/tex]

Justification: We rewrote [tex]\(-c\)[/tex] with a common denominator to match the denominator from the second term in the equation. This makes it possible to combine these terms into a single fraction:
[tex]\[ \frac{-4ac}{4a} + \frac{b^2}{4a} = \frac{-4ac + b^2}{4a} \][/tex]
Here, the justification is indeed the "common denominator."

For Step B:
[tex]\[ a\left(x+\frac{b}{2a}\right)^2 = \frac{-4ac + b^2}{4a} \][/tex]
[tex]\[ \left(\frac{1}{a}\right) a \left(x+\frac{b}{2a}\right)^2 = \left(\frac{1}{a}\right) \left(\frac{-4ac + b^2}{4a}\right) \][/tex]

Justification: We applied the multiplication property of equality to isolate [tex]\(\left(x+\frac{b}{2a}\right)^2\)[/tex]. Multiplying both sides of the equation by [tex]\(\frac{1}{a}\)[/tex] ensures that the equation remains balanced.

Thus, for Step B, the correct justification is the "multiplication property of equality."

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