# How do you derive the quadratic formula?

Feb 17, 2016

See explanation...

#### Explanation:

Given $a {x}^{2} + b x + c = 0$ to solve.

Note that:

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

So:

$0 = a {x}^{2} + b x + c = a {\left(x + \frac{b}{2 a}\right)}^{2} + \left(c - {b}^{2} / \left(4 a\right)\right)$

Add ${b}^{2} / \left(4 a\right) - c$ to both ends and transpose to get:

$a {\left(x + \frac{b}{2 a}\right)}^{2} = {b}^{2} / \left(4 a\right) - c = \frac{{b}^{2} - 4 a c}{4 a}$

Divide both sides by $a$ to get:

${\left(x + \frac{b}{2 a}\right)}^{2} = \frac{{b}^{2} - 4 a c}{4 {a}^{2}}$

Take the square root of both sides (allowing for either sign) to get:

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

Subtract $\frac{b}{2 a}$ from both sides to get:

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

Note that all of this is based on simple properties of arithmetic, so will work regardless of whether $a$, $b$ and $c$ are integers, rational, irrational or even Complex numbers.