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Difference between revisions of "2021 April MIMC 10 Problems/Problem 3"

(Created page with "Find the number of real solutions that satisfy the equation <math>(x^2+2x+2)^{3x+2}=1</math>. <math>\textbf{(A)} ~0 \qquad\textbf{(B)} ~1 \qquad\textbf{(C)} ~2 \qquad\textbf{...")
 
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==Solution==
 
==Solution==
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To be Released on April 26th.
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There are <math>2</math> cases when this expression can be equal to <math>1</math>: <math>x^2+2x+2=1</math> or <math>3x+2=0</math>. When <math>x^2+2x+2=1</math>, we can solve this quadratic to get <math>(x+1)^2=0</math>, or <math>x=-1</math>. We can solve the other solution by setting <math>3x+2=0</math>, or <math>x=-\frac{2}{3}</math>. However, we need to make sure that <math>x^2+2x+2\neq0</math> because <math>0^0</math> is undefined. Therefore, our answer would be <math>\fbox{\textbf{(C)} 2}</math>.

Latest revision as of 13:28, 26 April 2021

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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