How do you solve #k^2 + 6k - 24 = 0# by completing the square?

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Answer 1 · 2016-04-06T12:44:04

The solutions are:
#color(green)(k = sqrt 33 - 3# or , # color(green)(k = -sqrt 33 -3#

#k^2 + 6k - 24 = 0#

#k^2 + 6k = 24 #

To write the Left Hand Side as a Perfect Square, we add 9 to both sides
#k^2 + 6k + 9 = 24 + 9 #

#k^2 + 2 * k * 3 + 3^2 = 33 #

Using the Identity #color(blue)((a+b)^2 = a^2 + 2ab + b^2#, we get
#(k + 3)^2 = 33#

# k+ 3 = sqrt33# or # k + 3 = -sqrt33#

#color(green)(k = sqrt 33 - 3# or # color(green)(x = -sqrt 33 -3#