How do you solve #x^2+6x=7 # using completing the square?

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Answer 1 · 2015-06-19T22:14:37

#(x+3)^2 = x^2+6x+9#

So: #(x+3)^2 = 7 + 9 = 16 = 4^2#

Hence #x = -7# or #x = 1#

#(x+3)^2 = x^2+6x+9#

So: #(x+3)^2 = x^2+6x+9 = 7 + 9 = 16 = 4^2#

Take square root of both ends, allowing for both possibilities:

#x+3 = +-sqrt(4^2) = +-4#

Subtract #3# from both sides to get:

#x = -3+-4#

That is #x = -7# or #x = 1#