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Section 12.6 Derivation of the Quadratic Formula

The derivation of the quadratic equation relies on the process of completing the square. We being with the equation

\begin{equation*} ax^2+bx+c=0, a \neq 0\text{.} \end{equation*}

Before completing the square we need to isolate the constant and divide both sides by \(a\) so that the coefficient on the squared term is \(1\text{.}\)

\begin{align*} ax^2+bx+c\amp=0\\ ax^2+bx+c\subtractright{c}\amp=0\subtractright{c}\\ ax^2+bx\amp=-c\\ \multiplyleft{\frac{1}{a}}(ax^2+bx)\amp=\multiplyleft{\frac{1}{a}}-c\\ x^2+\frac{b}{a}x\amp=-\frac{c}{a} \end{align*}

We now complete the square on the left side of the equation by adding the square of half the linear coefficient. This action is balanced by adding the same expression to the right side of the equation. We'll then go ahead and combine the terms on the right side of the equation.

\begin{align*} x^2+\frac{b}{a}x\addright{\left(\frac{b}{2a}\right)^2}\amp=-\frac{c}{a}\addright{\left(\frac{b}{2a}\right)^2}\\ x^2+\frac{b}{a}x+\frac{b^2}{4a^2}\amp=-\frac{c}{a}+\frac{b^2}{4a^2}\\ x^2+\frac{b}{a}x+\frac{b^2}{4a^2}\amp=-\frac{c}{a} \highlight{\cdot \frac{4a}{4a}}+\frac{b^2}{4a^2}\\ x^2+\frac{b}{a}x+\frac{b^2}{4a^2}\amp=-\frac{4ac}{4a^2}+\frac{b^2}{4a^2}\\ x^2+\frac{b}{a}x+\frac{b^2}{4a^2}\amp=\frac{-4ac+b^2}{4a^2}\\ x^2+\frac{b}{a}x+\frac{b^2}{4a^2}\amp=\frac{b^2-4ac}{4a^2} \end{align*}

We now factor the left side into its perfect square and complete the derivation using the square root method. Note that when we simplify \(\sqrt{4a^2}\) we can simply state \(2a\text{.}\) It's true that we don't know whether \(a\) is a positive number or a negative number, but when we extract the roots we introduce the \(\pm\) sign, so we will have two roots, one positive and one negative, regardless of whether \(a\) is positive or negative (assuming that \(b^2-4ac \neq 0\)). If \(b^2-4ac=0\text{,}\) then \(\frac{b^2-4ac}{4a^2}=\frac{0}{2a}=0\) whether \(a\) is positive or negative.

\begin{align*} x^2+\frac{b}{a}x+\frac{b^2}{4a^2}\amp=\frac{b^2-4ac}{4a^2}\\ \left(x+\frac{b}{2a}\right)^2\amp=\frac{b^2-4ac}{4a^2}\\ x+\frac{b}{2a}\amp=\pm \sqrt{\frac{b^2-4ac}{4a^2}}\\ x+\frac{b}{2a}\amp=\pm \frac{\sqrt{b^2-4ac}}{2a}\\ x+\frac{b}{2a}\subtractright{\frac{b}{2a}}\amp=\pm \frac{\sqrt{b^2-4ac}}{2a}\subtractright{\frac{b}{2a}}\\ x\amp=-\frac{b}{2a}\pm \frac{\sqrt{b^2-4ac}}{2a}\\ x\amp=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \end{align*}
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