Proving the quadratic formula (for dummies)

Question

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.

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: http://math.stackexchange.com/questions/49229/why-can-all-quadratic-equations-be-solved-by-the-quadratic-formula/49243#49243 — Pedro M., May 14 '13 at 17:56

score 5 · Accepted Answer · answered May 14 '13 at 17:59

5

Look at each step here: $$ \begin{align*} a x^2 + b x + c &= 0 \\ a \left( x^2 + \frac{b}{a} x \right) + c &= 0 \\ a \left( x^2 + \frac{b}{a} x + \frac{b^2}{4 a^2} \right) - \frac{b^2}{4 a} + c &= 0 \\ a \left( x + \frac{b}{2 a} \right)^2 &= \frac{b^2}{4 a} - c \\ \left( x + \frac{b}{2 a} \right)^2 &= \frac{b^2 - 4 a c}{4 a^2} \\ x + \frac{b}{2 a} &= \frac{\pm\sqrt{b^2 - 4 a c}}{2 a} \\ x &= \frac{-b \pm\sqrt{b^2 - 4 a c}}{2 a} \end{align*} $$

answered May 14 '13 at 17:59

vonbrand

27,812

1

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. – kahen May 14 '13 at 18:10
thank you! Maths is not my strong point :) – imulsion May 15 '13 at 09:34

colormegone · Answer 2 · 2013-05-14T19:16:40.143

0

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?

edited May 14 '13 at 19:16

answered May 14 '13 at 17:53

colormegone

10,842

dreamer · Answer 3 · 2013-05-14T18:21:33.860

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.

Proving the quadratic formula (for dummies)

3 Answers3