1987 AHSME Problems/Problem 19

Problem

Which of the following is closest to $\sqrt{65}-\sqrt{63}$ ?

$\textbf{(A)}\ .12 \qquad \textbf{(B)}\ .13 \qquad \textbf{(C)}\ .14 \qquad \textbf{(D)}\ .15 \qquad \textbf{(E)}\ .16$

Solution

We have $\sqrt{65} > 8 > 7.5$ . Also $7.5^2 = (7 + 0.5)^2 = 7^2 + 2 \cdot 7 \cdot 0.5 + 0.5^2 = 49 + 7 + 0.25 = 56.25 < 63$ , so $\sqrt{63} > 7.5$ . Thus $\sqrt{65} + \sqrt{63} > 7.5 + 7.5 = 15$ . Now notice that $\sqrt{65} - \sqrt{63} = \frac{(\sqrt{65} - \sqrt{63})(\sqrt{65} + \sqrt{63})}{\sqrt{65} + \sqrt{63}} = \frac{2}{\sqrt{65} + \sqrt{63}$ (Error compiling LaTeX. Unknown error_msg), so $\sqrt{65} - \sqrt{63} < \frac{2}{15} = 0.1333333...$ , so the answer must be $A$ or $B$ . To determine which, we write $\sqrt{65} - \sqrt{63} > 0.125 \iff 65 - 2\sqrt{65 \cdot 63} + 63 > 0.015625 \iff 128 - 0.015625 > 2\sqrt{4095} \iff \sqrt{4095} < 64 - 0.0078125 \iff 4095 < 4096 - 32 \cdot 0.0078125 + 0.0078125^2 = 4096 - 0.25 + 0.0078125^2$ , which is true. Hence as the expression is greater than $0.125$ , and less than or equal to $0.13$ (since we showed it is certainly less than $0.1333333...$ ), it is closest to $0.13$ , which is answer $\boxed{\text{B}}$ .

1987 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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1987 AHSME Problems/Problem 19

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