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

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

1976 AHSME (Problems • Answer Key • Resources)
Preceded by 1975 AHSME	Followed by 1977 AHSME
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All AHSME Problems and Solutions

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 ==Solution==
-<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>.~MathJams
+{{AHSME box|year=1976|before=[[1975 AHSME]]|after=[[1977 AHSME]]}}

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

Latest revision as of 19:03, 12 July 2020

Problem 2

Solution