Quadratic Equation Solutions
Why
Given three nonzero real numbers $a, b, c$, find a real number $x$ to solve $ax^2 + bx + c = 0$.
Result
\[ \frac{-b + \sqrt{b^2 - 4ac}}{2a} \quad \text{ and } \quad \frac{-b - \sqrt{b^2 - 4ac}}{2a} \]
are solutions of\[ ax^2 + bx + c = 0. \]
We call $ax^2 + bx + c = 0$ a
quadratic equation.1
The solutions are often writen in short hand
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
\]
Proof
\[ x^2 + px + q = 0 \]
Then\[ x \in \set{-\frac{p}{2} + \sqrt{\frac{p^2}{4} - q}, -\frac{p}{2} - \sqrt{\frac{p^2}{4} - q}} \]
\[ x^2 + px + q = 0 \]
This holds if and only if\[ x^2 + px = -q \]
which holds if and only if\[ x^2 + px + \frac{p^2}{4} = \frac{p^2}{4} - q \]
We can express the left hand side as\[ x^2 + px + \frac{p^2}{4} = \left(x + \frac{p}{2}\right)^2 \]
By substituting this into the left hand side, we have that $x$ is a solution if and only if\[ \left(x + \frac{p}{2}\right)^2 = \frac{p^2}{4} - q, \]
Hence $x$ is a solution if and only if $x + p/2$ is a square root of $p^2/4 - q$. As usual, there are two quantities, which we denote as usual with $\sqrt{\cdot }$, so that $x$ is a solution if and only if\[ x + \frac{p}{2} = \pm\sqrt{\frac{p^2}{4} - q} \]
Consequently, $x$ is a solution if and only if\[ x = -\frac{p}{2} \pm \sqrt{\frac{p^2}{4} - q} \]
\[ x = -\frac{b}{2a} \pm \sqrt{\frac{b^2}{4a^2} - \frac{c}{a}} \]
Which holds if and only if\[ x = \frac{-b}{2a} \pm \sqrt{\frac{b^2}{4a^2} - \frac{4ac}{4a^2}} \]
which holds if and only if\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]
as desired.- Future editions will prove via completing the square. ↩︎