Question

Step A:

a (x + StartFraction b Over 2 a EndFraction) squared = –c + StartFraction b squared Over 2 a EndFraction

a (x + StartFraction b Over 2 a EndFraction) squared = StartFraction negative 4 a c + b squared Over 4 a EndFraction

Step B:

a (x + StartFraction b Over 2 a EndFraction) squared = StartFraction negative 4 a c + b squared Over 4 a EndFraction

(one-half) a (x + StartFraction b Over 2 a EndFraction) squared = (StartFraction 1 Over a EndFraction)(StartFraction b squared minus 4 a c Over 4 a EndFraction)

Determine the justification for the steps from the derivation of the quadratic formula.

Justification of step A:
Justification of step B:

Answer 1

Let us take a closer look at the very first steps; it will be easier to understand:

Step A.

$\quad a\left(x+\frac{b}{2a}\right)^{2}=-c+\frac{b^{2}}{2a}$

We simply do LCM of the RHS terms:

$\quad a\left(x+\frac{b}{2a}\right)^{2}=\frac{-2ac+b^{2}}{2a}$

In the question, it is given $\left(-c+\frac{b^{2}}{2a}\right)$ in the RHS, which is incorrect. It has to be $\left(-c+\frac{b^{2}}{4a}\right)$ there.

Step B.

Since this step is continuation of Step A, it is incorrect as well. Let us take the correct form.

Multiply both sides by $\frac{1}{a}$ with correct terms:

$\quad \frac{1}{a}.a.\left(x+\frac{b}{2a}\right)^{2}=\frac{1}{a}.\frac{-4ac+b^{2}}{4a}$

In the question, regardless mistakes in RHS, $\frac{1}{2}$ was multiplied in the LHS instead of $\frac{1}{a}$

With continuation for solution.

$\quad \left(x+\frac{b}{2a}\right)^{2}=\frac{b^{2}-4ac}{4a^{2}}$

$\Rightarrow x+\frac{b}{2a}=\pm\frac{\sqrt{b^{2}-4ac}}{2a}$

$\Rightarrow x=-\frac{b}{2a}\pm\frac{\sqrt{b^{2}-4ac}}{2a}$

$\Rightarrow \boxed{x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}}$

This is the required quadratic formula.

Answer 2

Answer:

Justification A: Common Denominator

Justification B: Multiplication Property of Equality