How do you solve #x^2 + 8x = 7# by completing the square?

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Answer 1 · 2016-05-09T22:12:46

#x = -4+-sqrt(23)#

If we add #16# to both sides then the left hand side of the equation becomes a perfect square:

#x^2+8x+16 = 23#

That is:

#(x+4)^2 = 23#

So #x+4# is a square root of #23#.

It could be the positive or negative one:

#x+4 = +-sqrt(23)#

Subtract #4# from both sides to find:

#x = -4+-sqrt(23)#