I have looked at this question, and also at this one, but I don't understand how the quadratic formula can change from $ax^2+bx+c=0$ to $x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$. I am not particularly good at maths, so can someone prove the quadratic formula in a simple way, with no complicated words? All help appreciated.
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2$\begingroup$ substitute and see what you get ! $\endgroup$– AmrMay 14, 2013 at 17:48
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$\begingroup$ @Amr what do you mean by that? $\endgroup$– imulsionMay 14, 2013 at 17:50
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$\begingroup$ math.stackexchange.com/a/176531/28743 $\endgroup$– nbubisMay 14, 2013 at 17:54
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$\begingroup$ Amr meant "Make $x$ equal to that formula and do the math". For a slightly different proof than usual, you might want to see this: math.stackexchange.com/questions/49229/… $\endgroup$– Pedro M.May 14, 2013 at 17:56
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1$\begingroup$ Completing the square works well. $\endgroup$– JulienMay 14, 2013 at 18:04
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3 Answers
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Look at each step here: $$ \begin{align*} a x^2 + b x + c &= 0 \\ a \left( x^2 + \frac{b}{a} x \right) + c &= 0 \\ a \left( x^2 + \frac{b}{a} x + \frac{b^2}{4 a^2} \right) - \frac{b^2}{4 a} + c &= 0 \\ a \left( x + \frac{b}{2 a} \right)^2 &= \frac{b^2}{4 a} - c \\ \left( x + \frac{b}{2 a} \right)^2 &= \frac{b^2 - 4 a c}{4 a^2} \\ x + \frac{b}{2 a} &= \frac{\pm\sqrt{b^2 - 4 a c}}{2 a} \\ x &= \frac{-b \pm\sqrt{b^2 - 4 a c}}{2 a} \end{align*} $$
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1$\begingroup$ Try multiplying through with $4a$ and then complete the square. Then you end up dividing by $2a$ only at the end and the rest of the manipulations are less messy. $\endgroup$– kahenMay 14, 2013 at 18:10
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$\begingroup$ thank you! Maths is not my strong point :) $\endgroup$– imulsionMay 15, 2013 at 9:34
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You should work this way to prove it: $ax^2+bx+c=0 \quad \iff \quad (\sqrt{a}x + \frac{1}{2}b/\sqrt{a})^2 + (c-\frac{1}{4}b^2/a) = 0$
$\iff \quad (\sqrt{a}x + \frac{1}{2}b/\sqrt{a})= + - \sqrt{\frac{1}{4}b^2/a -c}$
$\iff \quad \sqrt{a}x = -\frac{1}{2}b/\sqrt{a} + - \sqrt{\frac{1}{4}b^2/a -c} $
$\iff \quad x=-\frac{1}{2}b/a +- \sqrt{\frac{1}{4}b^2/a -c}/\sqrt{a}$
$\iff \quad x=-\frac{1}{2}b/a +- \sqrt{\frac{1}{4}b^2/a^2 -c/a}$
$\iff \quad \dots$
After some manipulations you will end up with the desired result.
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There are several ways to prove the quadratic formula. Using "completion of squares" is the simplest: rather than write it all out again, I can refer you to a post I wrote as part of a series of proofs.
Actually, though, it seems like a number of the links other people have pointed to have laid out the steps just as well. What specifically is the part that you don't understand?
answered May 14, 2013 at 17:53