If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Quadratic formula proof review

A text-based proof (not video) of the quadratic formula
The quadratic formula says that
x, equals, start fraction, minus, start color #e07d10, b, end color #e07d10, plus minus, square root of, start color #e07d10, b, end color #e07d10, squared, minus, 4, start color #7854ab, a, end color #7854ab, start color #e84d39, c, end color #e84d39, end square root, divided by, 2, start color #7854ab, a, end color #7854ab, end fraction
for any quadratic equation like:
start color #7854ab, a, end color #7854ab, x, squared, plus, start color #e07d10, b, end color #e07d10, x, plus, start color #e84d39, c, end color #e84d39, equals, 0
If you've never seen this formula proven before, you might like to watch a video proof, but if you're just reviewing or prefer a text-based proof, here it is:

The proof

We'll start with the general form of the equation and do a whole bunch of algebra to solve for x. At the heart of the proof is the technique called start color #11accd, start text, c, o, m, p, l, e, t, i, n, g, space, t, h, e, space, s, q, u, a, r, e, end text, end color #11accd. If you're unfamiliar with this technique, you may want to brush up by watching a video.

Part 1: Completing the square

ax2+bx+c=0(1)ax2+bx=c(2)x2+bax=ca(3)x2+bax+b24a2=b24a2ca(4)(x+b2a)2=b24a2ca(5)\begin{aligned} \purpleD{a}x^2 + \goldD{b}x + \redD{c} &= 0&(1)\\\\ ax^2+bx&=-c&(2)\\\\ x^2+\dfrac{b}{a}x&=-\dfrac{c}{a}&(3)\\\\ \blueD{x^2+\dfrac{b}{a}x+\dfrac{b^2}{4a^2}}&\blueD{=\dfrac{b^2}{4a^2}-\dfrac{c}{a}}&(4)\\\\ \blueD{\left (x+\dfrac{b}{2a}\right )^2}&\blueD{=\dfrac{b^2}{4a^2}-\dfrac{c}{a}}&(5) \end{aligned}

Part 2: Algebra! Algebra! Algebra!

Remember, our goal is to solve for x.
(x+b2a)2=b24a2ca(5)(x+b2a)2=b24a24ac4a2(6)(x+b2a)2=b24ac4a2(7)x+b2a=±b24ac4a2(8)x+b2a=±b24ac2a(9)x=b2a±b24ac2a(10)x=b±b24ac2a(11)\begin{aligned} \left (x+\dfrac{b}{2a}\right )^2&=\dfrac{b^2}{4a^2}-\dfrac{c}{a}&(5) \\\\ \left (x+\dfrac{b}{2a}\right )^2&=\dfrac{b^2}{4a^2}-\dfrac{4ac}{4a^2} &(6)\\\\ \left (x+\dfrac{b}{2a}\right )^2&=\dfrac{b^2-4ac}{4a^2}&(7)\\\\ x+\dfrac{b}{2a}&=\pm \dfrac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}&(8)\\\\ x+\dfrac{b}{2a}&=\pm \dfrac{\sqrt{b^2-4ac}}{2a}&(9)\\\\ x&=-\dfrac{b}{2a}\pm \dfrac{\sqrt{b^2-4ac}}{2a}&(10)\\\\ x&=\dfrac{-\goldD{b}\pm\sqrt{\goldD{b}^2-4\purpleD{a}\redD{c}}}{2\purpleD{a}}&(11) \end{aligned}
And we're done!

Want to join the conversation?