How did they find the Quadratic Formula?

A few weeks ago, we were looking at quadratic equations, equations with an x^2 term in it. We learned how to complete the square, which is translating from standard form to vertex form.

Another cool thing I never mentioned that is cool about vertex form is that you can calculate x when the equation is in vertex form. For (x + 3)^2 – 1 = 0, you can say that x = -2 or -4, just by simply manipulating variables.

(x + 3)^2 – 1 = 0

(x + 3)^2 = 1                 (add one to both sides)

x + 3 = 1 or -1               (take the ± square root of both sides; use 1 and -1 because 1
                                      and -1 squared are both 1)

x = -2 or -4                    (subtract three from both sides, from the 1 and the -1 to get
                                      both answers)

What about for standard form, using the coefficients a, b, and c. Maybe there is a formula we can derive just by following the steps involved in completing the square. Remember our original equation: ax² + bx + c = 0. We’ll do it alongside an example utilizing every step, 2x² – 4x + 6 = 0.

Starting Equation	ax2 + bx + c = 0	2x2 – 4x – 6 = 0
Move the constant over to the right hand side, just to make things less messy.	ax2 + bx = -c	2x2 – 4x = 6
Divide through by a. Normally, we would factor it, but it is the same thing.	x2 + b/ax = -c/a	x2 – 2x = 3
Take half of the new b term and square it.	(b/2a)2 = b^2/4a^2	(-1)2 = 1
Add in and subtract out this number. For this, we’ll just add it to both sides, since c is on the other side.	x2 + b/ax + b^2/4a^2 = -c/a + b^2/4a^2	x2 – 2x + 1 = 3 + 1
Simplify the right hand side. For the first example, we will multiply the first fraction by 4a/4a to get a common denominator.	4a/4a(-c/a) + b^2/4a^2 -4ac/4a^2 + b^2/4a^2 b^2 – 4ac/4a^2 x2 + b/ax + b^2/4a^2 = b^2 – 4ac/4a^2	x2 – 2x + 1 = 4
Convert the left hand side into squared form.	(x + b/2a)2 = b^2 – 4ac/4a^2	(x – 1)2 = 4
Square root both sides. Don't forget to put our ± sign before the square root!	x + b/2a = ±√(b^2 – 4ac)/2a	x – 1 = ±2
Isolate the x by adding/subtracting the remaining constant.	x = -b/2a ± √(b^2 – 4ac)/2a x = -b ± √(b^2 – 4ac)/2a	x = 1 ± 2 x = -1 or 3

And we have a formula; x = -b ± √(b^2 – 4ac)/2a. This is known as the quadratic formula, the formula that takes ANY quadratic equation and generates its x-intercepts. Not only that, but it doesn't generate false intercepts, and it is very clear when there are no intercepts.

This formula is pretty well-known, but I think it's cool how it is derived. This is a formula that you want to memorize, because it is so important.