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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2021 April MIMC 10 Problems/Problem 3

Solution