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

1 Answer
Apr 23, 2016

Answer:

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

Explanation:

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#