Note: This is not needed for exams. It requires moderate
algebra knowledge and uses completing the square.
How to derive the quadratic formula
Start with the general quadratic equation and make \(x\) the subject:
\[0=ax^2+bx+c\]
Factor \(a\) out and divide through on both sides:
\[\begin{align}
0&=a\left(x^2+\frac{bx}{a}+\frac{c}{a}\right)
\\[8pt]0&=x^2+\frac{bx}{a}+\frac{c}{a}
\end{align}\]
Complete the square,
not expanding \((2a)^2\) in some cases to avoid later factorisation:
\[\begin{align}
x^2+\frac{bx}{a}+\frac{c}{a}&=\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{(2a)^2}+\frac{c}{a}
\\[8pt]0&=\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{(2a)^2}+\frac{c}{a}
\\[8pt]\left(x+\frac{b}{2a}\right)^2&=\frac{b^2}{(2a)^2}-\frac{c}{a}
\\[8pt]x+\frac{b}{2a}&=\pm\sqrt{\frac{b^2}{(2a)^2}-\frac{c}{a}}
\\[8pt]x&=\frac{-b}{2a}\pm\sqrt{\frac{b^2}{(2a)^2}-\frac{c}{a}}
\end{align}\]
This would work as an alternative but let's rearrange to make it easier to use:
\[\begin{align}
\frac{-b}{2a}\pm\sqrt{\frac{b^2a}{(2a)^2a}-\frac{(2a)^2c}{(2a)^2a}}&=x
\\[8pt]\frac{-b}{2a}\pm\sqrt{\frac{b^2a-4a^2c}{(2a)^2a}}&=x
\\[8pt]\frac{-b}{2a}\pm\sqrt{\frac{b^2-4ac}{(2a)^2}}&=x
\\[8pt]\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{\sqrt{(2a)^2}}&=x
\\[8pt]\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}&=x
\\[8pt]\frac{-b\pm\sqrt{b^2-4ac}}{2a}&=x
\end{align}\]
There you have it! Obviously this is fairly long-winded to do during an exam so you should learn the last equation by heart. However, it's always good to know where things come from.
