How do you solve #x^2-14x+19=0 # by completing the square?

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Answer 1 · 2016-04-18T17:15:06

The solutions are:
#color(green)(x = sqrt 30 + 7# ,# color(green)(x = -sqrt 30 + 7#

#x^2 - 14x + 19 =0 #

#x^2 - 14x = - 19#

To write the Left Hand Side as a Perfect Square, we add 49 to both sides:

#x^2 - 14x + 49 = - 19 + 49#

#x^2 - 2 * x * 7 + 7^2 = 30#

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

#(x-7)^2 = 30#

#x - 7 = sqrt30# or #x - 7 = -sqrt30#

#color(green)(x = sqrt 30 + 7# or # color(green)(x = -sqrt 30 + 7#