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Difference between revisions of "1958 AHSME Problems/Problem 5"

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<math>\frac{1}{2 + \sqrt2} \cdot \frac{2 - \sqrt2}{2 - \sqrt2} \Rightarrow \frac{2 - \sqrt2}{2}</math>.
 
<math>\frac{1}{2 + \sqrt2} \cdot \frac{2 - \sqrt2}{2 - \sqrt2} \Rightarrow \frac{2 - \sqrt2}{2}</math>.
  
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Now, the other fraction we need to get the radical out of the denominator is <math>\frac{1}{\sqrt2 - 2}</math>. Here, were will also multiply by the conjugate, <math>\sqrt2 + 2</math>. So that simplifies to
+
Now, the other fraction we need to get the radical out of the denominator is <math>\frac{1}{\sqrt2 - 2}</math>. Here, we will multiply by the conjugate again, <math>\sqrt2 + 2</math>. So that simplifies to
  
 
<math>\frac{1}{\sqrt2 - 2} \cdot \frac{\sqrt2 + 2}{\sqrt2 + 2} \Rightarrow \frac{\sqrt2 + 2}{-2}</math>.
 
<math>\frac{1}{\sqrt2 - 2} \cdot \frac{\sqrt2 + 2}{\sqrt2 + 2} \Rightarrow \frac{\sqrt2 + 2}{-2}</math>.

Revision as of 02:00, 22 November 2016

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

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

See also

1958 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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All AHSME Problems and Solutions

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