Difference between revisions of "1965 AHSME Problems/Problem 11"
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== Solution == | == Solution == | ||
− | <math>\fbox{B}</math> | + | |
+ | Statement I is incorrect, because <math>(\sqrt{-4})(\sqrt{-16})=(i\sqrt{4})(i\sqrt{16})=i^2(\sqrt{4})(\sqrt{16})=-\sqrt{64}</math>. <math>\newline</math> | ||
+ | Statement II is correct, because <math>(-4)(-16)=64</math>, so <math>\sqrt{(-4)(-16)}=\sqrt{64}</math>. <math>\newline</math> | ||
+ | Statement III is correct, because <math>8^2=64</math> and <math>8 \geq 0</math> (so it is the [[square root|principal square root]] of 64). <math>\newline</math> | ||
+ | Thus, only statement I is incorrect, so we choose answer <math>\fbox{B}</math>. | ||
==See Also== | ==See Also== | ||
− | {{AHSME 40p box|year=1965|num-b= | + | {{AHSME 40p box|year=1965|num-b=10|num-a=12}} |
+ | {{MAA Notice}} |
Latest revision as of 16:02, 18 July 2024
Problem
Consider the statements:
Solution
Statement I is incorrect, because . Statement II is correct, because , so . Statement III is correct, because and (so it is the principal square root of 64). Thus, only statement I is incorrect, so we choose answer .
See Also
1965 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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All AHSME Problems and Solutions |
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