2021 April MIMC 10 Problems/Problem 3

Find the number of real solutions that satisfy the equation $(x^2+2x+2)^{3x+2}=1$.

$\textbf{(A)} ~0 \qquad\textbf{(B)} ~1 \qquad\textbf{(C)} ~2 \qquad\textbf{(D)} ~3 \qquad\textbf{(E)} ~4$

Solution

There are $2$ cases when this expression can be equal to $1$: $x^2+2x+2=1$ or $3x+2=0$. When $x^2+2x+2=1$, we can solve this quadratic to get $(x+1)^2=0$, or $x=-1$. We can solve the other solution by setting $3x+2=0$, or $x=-\frac{2}{3}$. However, we need to make sure that $x^2+2x+2\neq0$ because $0^0$ is undefined. Therefore, our answer would be $\fbox{\textbf{(C)} 2}$.

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