1976 AHSME Problems/Problem 2

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Problem 2

For how many real numbers $x$ is $\sqrt{-(x+1)^2}$ a real number?

$\textbf{(A) }\text{none}\qquad \textbf{(B) }\text{one}\qquad \textbf{(C) }\text{two}\qquad\\ \textbf{(D) }\text{a finite number greater than two}\qquad \textbf{(E) }\infty$

Solution

$\sqrt{-(x+1)^2}$ is a real number, if and only if $-(x+1)^2$ is nonnegative. Since $(x+1)^2$ is always nonnegative, $-(x+1)^2$ is nonnegative only when $-(x+1)^2=0$, or when $x=-1 \Rightarrow \textbf{(B)}$.~MathJams


1976 AHSME (ProblemsAnswer KeyResources)
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