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.

Answer 1

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

Answer 2

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?

Answer 3

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.