How do you solve the equation #x^2+7x+15=0# by completing the square?

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Answer 1 · 2017-04-05T15:50:45

#(x + 7/2)^2 + 11/4 = 0#

The process of completing the square yields the standard form or
Vertex form #y = a(x-h)^2 + k#, where vertex: #(h, k)#

Use completing of the square by grouping the #x#-terms:
#(x^2 + 7x) +15 = 0#

Get the squared value by halving the #x#-term: #1/2 *7 = 7/2#
#(x +7/2)^2 +15 - (7/2)^2 = 0#

You need to subtract #(7/2)^2 = 49/4# because it is added when the square is completed:

#(x+7/2)^2 = (x+7/2)(x+7/2) = x^2 +7x + 49/4#

From #" " (x + 7/2)^2 +15 - 49/4 = 0#

#" " (x + 7/2)^2 +60/4 - 49/4 = 0#

#(x+7/2)^2 + 11/4 = 0#