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A text-based proof (not video) of the quadratic formula
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
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

\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.
\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?

• When we take the sqrt of 4a^2 shouldn't it be + or - 2a, not just 2a?
• anytime you take the square root of an expression, it is considered the principal root - that means the value is only ever positive... when you have an equation and take the square root of both sides, that is when you can get two values... (you can get as many values as makes the equation true) ex:
x^2 = 4
√(x^2) = √4
|x| = 2
x = ±2
• In step 8 the square root on the right hand side is +/-. Why is the square root on the left hand side not also +/-?
• on the left hand side x + (b/2a) is squared so the root cancels out
• How can we multiply by 4a^2 in step 6, without affecting the left side of the equation?
• What they did in step 6 was multiply - c/a by 4a/4a. The reason you can do this is because 4a/4a is the same thing as 1, so multiplying by it doesn't change any values.
In other words, -c/a has the same value as -4ac/(4a^2)
• number one victory royale...
• I require some help with understanding how -b/2a derives the x-coordinate of the vertex of a parabola. Thanks in advance!
(1 vote)
• I know of two ways to understanding it.

First, using the vertex formula: y = a(x – h)^2 + k, where "h" is the vertex.
Put the general equation y = ax^2 + bx + c into the vertex form and you will find that "h" will equal -b/2a. I'll leave the work up to you.

Second, since quadratics in the general form (y = ax^2 + bx + c) are symmetric over a vertical line through the vertex, we can use the two roots of the quadratic formula and average them to find the x-coordinate of the vertex (visualize a quadratic graph and you will see why this is true).

So if you find the average of the two roots:

[-b + sqrt(b^2-4ac)]/2a and [-b - sqrt(b^2-4ac)]/2a // it will be -b/2a. (I again, will leave the work up to you.)
• n step 8 the square root on the right hand side is +/-. Why is the square root on the left hand side not also +/-?
• That is a great question! I thought I knew algebra, but I never noticed that and it took me a little minute to work out!
Only one side of the equation needs a +/- sign because if you multiply the equation by -1 you can get to any combination of negatives and positives while only putting the +/- sign on one side. I know that sounds confusing, so let's simplify things. Say, for example, we're using the equation a = ±b. The two possible situations are a = b and a = -b, but multiplying those by -1 gives you -a = -b and -a = b, meaning that either side could be positive or negative in any combination with the +/- sign on only one side. Terrific question! I hope this helps, and remember that you can learn anything!
• can someone help me with 4x^2+11x-20=0 I solved everything expect I got stuck on the square root of 441
• why don't we write +/- when we take the square root of 4a^2 ?
(1 vote)
• Look at answer to similar question 8 days ago from emilio12medina. The idea is that you do not need two sets of ± signs, they end up cancelling each other out.
• I tried the proof myself in a slightly different way and it didn't quite work out.

(1) ax^2 + bx + c = 0
(2) x^2 + (b/a)x + c/a = 0
(3) x^2 + (b/a)x + (b^2/4a^2) + c/a - (b^2/4a^2) = 0
(4) (x+(b/2a))^2 + 4ac/4a^2 - b^2/4a^2 = 0
(5) (x+(b/2a))^2 + (4ac-b^2)/(4a^2) = 0
(6) (x+(b/2a))^2 = -(4ac-b^2)/(4a^2)
(7) (x+(b/2a)) = -(sqrt(4ac-b^2))/2a
(8) x = -(b/2a)-(sqrt(4ac-b^2))/2a
(9) x = -(-b+-squr(4ac-b^2))/2a

My discriminant is 4ac-b^2 while the one we use is b^2-4ac. Also, I have a negative sign in front of the whole fraction which is from the second equation in step 8. Could somebody tell me where I made a mistake?