## College Algebra (10th Edition)

The solution set is $\left\{-1\right\}$.
Raise both sides of the equation to the 5th power to obtain: $x^2+2x = (-1)^5 \\x^2+2x=-1$ Add 1 on both sides of the equation to obtain: $x^2+2x+1=0$ With $a=1, b=2, c=1$, use the quadratic formula $x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ to obtain: $x=\dfrac{-2 \pm \sqrt{2^2-4(1)(1)}}{2(1)} \\x=\dfrac{-2\pm\sqrt{4-4}}{2} \\x=\dfrac{-2\pm0}{2} \\x=-1$ Thus, the solution set is $\left\{-1\right\}$.