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

1 Answer
Stefan V.
Aug 12, 2015

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)):}

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