1965 AHSME Problems/Problem 11

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.

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

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, and so we choose answer $\fbox{B}$ .

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

Page

Toolbox

Search

1965 AHSME Problems/Problem 11

Problem

Solution

See Also