How do you solve #x^2-12x=-28# by completing the square?

1 Answer
Sep 3, 2016

Answer:

#x =8.828 or x = 3.172#

Explanation:

#(x+- p)^2 = x^2 +-2px + p^2#

If you have the correct combination of 3 terms, you can write the expression as #(x+-p)^2#
Add the correct last term each time, to 'complete the square.'

#x^2 color(blue)(-12)x + square = -28 " "square = (color(blue)(-12)/2)^2 = color(red)(36)#
#x^2 color(blue)(-12)x + color(red)(36) = -28 +color(red)(36)#

#(x-6)^2 = 8 " " -6 = color(blue)(-12)/2 #

#x- 6 = +-sqrt8#

#x = sqrt8+6 =8.828 " or " x = -sqrt8+6 =3.172 #