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Difference between revisions of "1976 AHSME Problems/Problem 2"

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<math>\sqrt{-(x+1)^2}</math> is a real number, if and only if <math>-(x+1)^2</math> is nonnegative. Since <math>(x+1)^2</math> is always nonnegative, <math>-(x+1)^2</math> is nonnegative only when <math>-(x+1)^2=0</math>, or when <math>x=-1 \Rightarrow \textbf{(B)}</math>.
 
<math>\sqrt{-(x+1)^2}</math> is a real number, if and only if <math>-(x+1)^2</math> is nonnegative. Since <math>(x+1)^2</math> is always nonnegative, <math>-(x+1)^2</math> is nonnegative only when <math>-(x+1)^2=0</math>, or when <math>x=-1 \Rightarrow \textbf{(B)}</math>.
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==1976 AHSME Problems==
 
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{{AHSME box|year=1976|before=[[1975 AHSME]]|after=[[1977 AHSME]]}}
 

Revision as of 19:03, 12 July 2020

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 $x=-1 \Rightarrow \textbf{(B)}$.

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