1965 AHSME Problems/Problem 11

Problem

Consider the statements: I:(4)(16)=(4)(16),II:(4)(16)=64,III:64=8. Of these the following are incorrect.

$\textbf{(A)}\ \text{none} \qquad  \textbf{(B) }\ \text{I only} \qquad  \textbf{(C) }\ \text{II only} \qquad  \textbf{(D) }\ \text{III only}\qquad \textbf{(E) }\ \text{I and III only}$

Solution

Statement I is incorrect, because $(\sqrt{-4})(\sqrt{-16})=(i\sqrt{4})(i\sqrt{16})=i^2(\sqrt{4})(\sqrt{16})=-\sqrt{64}$. $\newline$ Statement II is correct, because $(-4)(-16)=64$, so $\sqrt{(-4)(-16)}=\sqrt{64}$. $\newline$ Statement III is correct, because $8^2=64$ and $8 \geq 0$ (so it is the principal square root of 64). $\newline$ Thus, only statement I is incorrect, so we choose answer $\fbox{B}$.

See Also

1965 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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