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=10|num-a=12}} | {{AHSME 40p box|year=1965|num-b=10|num-a=12}} | ||
Latest revision as of 11:19, 18 July 2024
Problem
Consider the statements: \begin{align} &I: (\sqrt{-4})(\sqrt {-16}) = \sqrt{(-4)(-16)}, \\ &II: \sqrt{(-4)(-16)} = \sqrt{64}, \\ &III: \sqrt{64} = 8. \end{align} Of these the following are incorrect.
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 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
