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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?
• When you took the square root of the entire fraction, technically both sides have the 'plus or minus' symbol but putting it on both the numerator and the denominator would be redundant and you can simply simplify it like they did with a single plus or minus symbol in front of the fraction but you can also leave it on the numerator.
• 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)
• 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.)
• can someone help me with 4x^2+11x-20=0 I solved everything expect I got stuck on the square root of 441
• 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?
• At step 6 and 7 when you took the square root, the negative should have stayed inside rather than outside (taking square root would yield ± on the outside). So when you distribute the -1(4ac-b^2) you end up with b^2-4ac.
(1 vote)
• on step 4, why adding (b^2/4a^2) to both sides?
(1 vote)
• To complete the square, you divide the coefficient of the x term by 2 (b/2a) and square this to get b^2/4a^2. So you need this term to complete the square. If you do it to the left side in order to complete the square, you either have to subtract it on the left or add it to the right side of the equation to keep it balanced.