How do you derive the quadratic formula?

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How do you derive the quadratic formula?

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Answer 1 · 2015-01-10T22:03:09

This is a little bit tricky but also incredibly elegant!
You start from your general quadratic:
$a x^{2} + b x + c = 0$ $ax^2+bx+c=0$
Take $c$ $c$ to the right side:
$a x^{2} + b x = - c$ $ax^2+bx=-c$
The idea is now to transform the left side in something like ${(a + b)}^{2}$ $(a+b)^2$ ;
Multiply by $a$ $a$ ;
$a^{2} x^{2} + a b x = - a c$ $a^2x^2+abx=-ac$
Multiply by $4$ $4$ ;
$4 a^{2} x^{2} + 4 a b x = - 4 a c$ $4a^2x^2+4abx=-4ac$
Add and subtract $b^{2}$ $b^2$ to the left side:
$4 a^{2} x^{2} + 4 a b x + b^{2} - b^{2} = - 4 a c$ $4a^2x^2+4abx+b^2-b^2=-4ac$
Take the $- b^{2}$ $-b^2$ to the right:
$4 a^{2} x^{2} + 4 a b x + b^{2} = b^{2} - 4 a c$ $4a^2x^2+4abx+b^2=b^2-4ac$
The left side can be written as:
${(2 a x + b)}^{2} = b^{2} - 4 a c$ $(2ax+b)^2=b^2-4ac$
And:
$2 a x + b = \pm \sqrt{b^{2} - 4 a c}$ $2ax+b=+-sqrt(b^2-4ac)$
And finally:
$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$
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How do you derive the quadratic formula?

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