1958 AHSME Problems/Problem 5

Problem

The expression $2 + \sqrt{2} + \frac{1}{2 + \sqrt{2}} + \frac{1}{\sqrt{2} - 2}$ equals:

$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 2 - \sqrt{2}\qquad \textbf{(C)}\ 2 + \sqrt{2}\qquad \textbf{(D)}\ 2\sqrt{2}\qquad \textbf{(E)}\ \frac{\sqrt{2}}{2}$

Solution

To make this problem easier to solve, lets get the radicals out of the denominator. For $\frac{1}{2 + \sqrt2}$ , we will multiply the numerator and denominator by $2- \sqrt2$ so,

$\frac{1}{2 + \sqrt2} \cdot \frac{2 - \sqrt2}{2 - \sqrt2} \Rightarrow \frac{2 - \sqrt2}{2}$ .

Now, the other fraction we need to get the radical out of the denominator is $\frac{1}{\sqrt2 - 2}$ . Here, we will multiply by the conjugate again, $\sqrt2 + 2$ . So that simplifies to

$\frac{1}{\sqrt2 - 2} \cdot \frac{\sqrt2 + 2}{\sqrt2 + 2} \Rightarrow \frac{\sqrt2 + 2}{-2}$ .

So now our simplified equation is

$2 + \sqrt2 + \frac{2- \sqrt2}{2} + \frac{\sqrt2 + 2}{-2} \Rightarrow 2+ \sqrt2 + \frac{2 - \sqrt2}{2} - \frac{\sqrt2 +2}{2}$

Bringing everything to the same denominator and combining like terms, we get

$\frac{4 + 2\sqrt2 + 2 - \sqrt2 - \sqrt2 - 2}{2} \Rightarrow \frac{4}{2} \Rightarrow 2 \Rightarrow \boxed{A}$

1958 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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1958 AHSME Problems/Problem 5

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Solution

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