Question

Consider the first five steps of the derivation of the Quadratic Formula.

Answer 1

We want to derive the quadratic formula from the following expression

$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}$

The next step would be rewrite the left side of the equation as the square of a sum, and combine the two terms on the right side of the equation on a single fraction.

Let's start by rewriting the left side.

When we expand the square of a binomial, we have

$(m+n)^2=m^2+2mn+n^2$

In our expression, we have

$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=x^2+\frac{b}{a}x+(\frac{b}{2a})^2$

Comparing our expression with the expansion of a squared binomial, we have

$\begin{cases}m=x \\ n=\frac{b}{2a}\end{cases}\Rightarrow x^2+\frac{b}{a}x+(\frac{b}{2a})^2=(x+\frac{b}{2a})^2$

Then, our original expression can be rewritten as

$\begin{gathered} x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2} \\ (x+\frac{b}{2a})^2=-\frac{c}{a}+\frac{b^2}{4a^2} \end{gathered}$

And finally, by combining the terms on the right side, we have

$\begin{gathered} (x+\frac{b}{2a})^2=-\frac{c}{a}+\frac{b^2}{4a^2} \\ (x+\frac{b}{2a})^2=-\frac{c}{a}\cdot\frac{4a}{4a}+\frac{b^2}{4a^2} \\ (x+\frac{b}{2a})^2=-\frac{4ac}{4a^2}+\frac{b^2}{4a^2} \\ (x+\frac{b}{2a})^2=\frac{-4ac+b^2}{4a^2} \\ (x+\frac{b}{2a})^2=\frac{b^2-4ac^{}}{4a^2} \end{gathered}$

This is the next step. The answer is option C.