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

1 Answer
Aug 12, 2015

Answer:

#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)):}#