How do you derive the quadratic formula?

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How do you derive the quadratic formula?

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Answer 1 · 2016-02-17T22:15:27

See explanation...

Explanation:

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

Note that:

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

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:

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

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

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

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

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

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

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

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

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How do you derive the quadratic formula?

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