How do you solve #x^2 + 4x - 12 = 0# by completing the square?

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Answer 1 · 2016-05-20T11:58:13

The solutions are #color(green)(x = 2# ,# color(green)(x = -6#

#x^2 + 4x - 12 = 0 #

#x^2 + 4x = 12#

To write the Left Hand Side as a Perfect Square, we add 4 to both sides:

#x^2 + 4x + 4 = 12 + 4#

#x^2 + 2 * x * 2 + 2^2 = 16#

Using the Identity #color(blue)((a+b)^2 = a^2 + 2ab + b^2#, we get
#(x+2)^2 = 16#

#x + 2 = sqrt16# or #x +2 = -sqrt16#

#x + 2 = 4# or #x +2 = -4#

#x = 4 -2 # or #x = -4 -2 #

#color(green)(x = 2# ,# color(green)(x = -6#