Proof: The quadratic formula
Before reading this proof sheet, it is recommended that you read Guide: Completing the square.
Proof of the quadratic formula
Remember from Guide: Using the quadratic formula that the quadratic formula is used to find roots of any quadratic equation:
Let \(ax^2 + bx + c = 0\) be a quadratic equation (where \(a\neq 0\)). The roots to this quadratic equation are given by \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] where one of the roots is given by the term \((-b+\sqrt{b^2 - 4ac})/2a\) and the other given by the term \((-b-\sqrt{b^2 - 4ac})/2a\).
In order to prove that these really are the solutions to the quadratic, you can complete the square on \(ax^2 + bx + c\) using the fact that \(a\neq 0\). See Guide: Completing the square for why this works.
Proof of the quadratic formula
First of all, as \(a\neq 0\) you can divide \(ax^2 + bx + c = 0\) through by \(a\) to get \[x^2 + \frac{b}{a} x + \frac{c}{a} = 0\] Taking the \(c/a\) term over to the other side gives \[x^2 + \frac{b}{a} x = -\frac{c}{a}\] Completing the square (see Guide: Completing the square) gives \[\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} = -\frac{c}{a}\] You can rearrange to get \[\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{c}{a} = \frac{b^2 - 4ac}{4a^2}\] Now the result is starting to come together. Taking square roots of both sides (not forgetting that it could be positive or negative) gives \[x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}\] and rearranging gives \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] as required.
Further reading
Guide: Using the quadratic formula
Questions: Using the quadratic formula
Version history and licensing
v1.0: created in 04/24 by tdhc.