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:

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

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