Question

How do you solve `x^2-12x+20=0` by completing the square?

Answer 1

`x_(1,2) = 6 +- 4`

Explanation:

Start by getting your quadratic into the form

`color(blue)(x^2 + b/ax = -c/a)`

This can be done by adding `-20` to both sides of the equation

`x^2 - 12x + color(red)(cancel(color(black)(20))) - color(red)(cancel(color(black)(20))) = 0 - 20`

`x^2 - 12x = -20`

Now use the coefficient of the `x`-term to find a term that, when added to both sides of the equation, will allow you to write the left side as the square of a binomial.

More specifically, you need to divide this coefficient by `2` and square the result.

`((-12)/2)^2 = (-6)^2 = 36`

So, add `36` to both sides of the equation to get

`x^2 - 12x + 36 = -20 + 36`

The left side of the equation can now be written as

`x^2 - 2 * (6) * x + (6)^2 = (x-6)^2`

This means that you have

`(x-6)^2 = 16`

Take the square root of both sides of the equation to get

`sqrt((x-6)^2) = sqrt(16)`

`x-6 = +- 4`

`x = 6 +- 4 = {(x_1 = 6 + 4 = color(green)(10)), (x_2 = 6 - 4 = color(green)(2)):}`