Certainly! Let's break down the two steps and their justifications for deriving the quadratic formula.
### Step A:
Given equation:
[tex]\[ a\left(x+\frac{b}{2a}\right)^2 = -c + \frac{b^2}{4a} \][/tex]
To combine the terms on the right-hand side into a single fraction with a common denominator:
[tex]\[ a\left(x+\frac{b}{2a}\right)^2 = \frac{-4ac + b^2}{4a} \][/tex]
Justification for Step A:
- Common denominator: We combined the terms [tex]\(-c\)[/tex] and [tex]\(\frac{b^2}{4a}\)[/tex] by expressing [tex]\(-c\)[/tex] as [tex]\(\frac{-4ac}{4a}\)[/tex]. This allows us to add the fractions together, yielding the single fraction [tex]\(\frac{-4ac + b^2}{4a}\)[/tex].
### Step B:
Continuing from:
[tex]\[ a\left(x+\frac{b}{2a}\right)^2 = \frac{-4ac + b^2}{4a} \][/tex]
Multiply both sides by the reciprocal of [tex]\(a\)[/tex] to solve for [tex]\( \left(x+\frac{b}{2a}\right)^2 \)[/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]
[tex]\[ \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} \][/tex]
Justification for Step B:
1. Multiplication property of equality: We multiplied both sides of the equation by [tex]\(\frac{1}{a}\)[/tex] to isolate [tex]\( \left(x + \frac{b}{2a}\right)^2 \)[/tex] on the left-hand side.
2. Common denominator: This multiplication also results in a further simplification because the factor of [tex]\(a\)[/tex] cancels out in the denominator on the right-hand side, conforming to the standard algebraic steps of combining and simplifying fractions.
Therefore, the justifications for step B are:
- Multiplication property of equality
- Common denominator
The final conclusion of the justifications for the steps is as follows:
- Step A: Common denominator
- Step B: Multiplication property of equality, Common denominator
