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>.~MathJams |
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{{AHSME box|year=1976|before=[[1975 AHSME]]|after=[[1977 AHSME]]}} | {{AHSME box|year=1976|before=[[1975 AHSME]]|after=[[1977 AHSME]]}} | ||
Latest revision as of 19:03, 12 July 2020
Problem 2
For how many real numbers 
is 
a real number?
Solution
is a real number, if and only if 
is nonnegative. Since 
is always nonnegative, is nonnegative only when 
, or when 
.~MathJams
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