What is the quadratic formula?

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What is the quadratic formula?

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Answer 1 · 2017-03-02T19:13:35

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

Explanation:

Given a quadratic equation in the standard form:

$a x^{2} + b x + c = 0$ $ax^2+bx+c = 0$

the roots are given by the quadratic formula:

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

This is a very useful formula to memorise, but where does it come from?

Given:

$a x^{2} + b x + c = 0$ $ax^2+bx+c = 0$

Note that:

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

$a {(x + \frac{b}{2 a})}^{2} = a x^{2} + b x + \frac{b^{2}}{4 a}$ $color(white)(a(x+b/(2a))^2) = ax^2+bx+b^2/(4a)$

So:

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

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

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

$a {(x + \frac{b}{2 a})}^{2} = \frac{b^{2} - 4 a c}{4 a}$ $color(white)(a(x+b/(2a))^2) = (b^2-4ac)/(4a)$

Divide both sides by $a$ $a$ to get:

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

${(x + \frac{b}{2 a})}^{2} = \frac{b^{2} - 4 a c}{{(2 a)}^{2}}$ $color(white)((x+b/(2a))^2) = (b^2-4ac)/(2a)^2$

Take the square root of both sides, allowing for both positive and negative square roots:

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

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

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

Note that the quadratic formula will always work, but it will only give real values if $b^{2} - 4 a c \geq 0$ $b^2-4ac >= 0$ , resulting in real square roots.

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What is the quadratic formula?

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