How do you solve #x^2+6x+9=-32#?

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Answer 1 · 2016-04-23T23:17:55

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

The left hand side here is a perfect square trinomial, but the right hand side is negative, so only has an imaginary square root.

We find:

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

Hence:

#(x+3)^2 = -32#

So:

#x+3 = +-sqrt(-32) = +-sqrt(32)i = +-4sqrt(2)i#

Subtract #3# from both sides to get:

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