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

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==Solution==
 
==Solution==
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We have
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<cmath>\begin{align*}2+\sqrt{2}+\frac{2-\sqrt{2}}{4-2}+\frac{\sqrt{2}+2}{2-4}&=2+\sqrt{2}+\frac{1}{2}(2-\sqrt{2})-\frac{1}{2}(\sqrt{2}+2)\\&=\frac{1}{2}(2+\sqrt{2}+2-\sqrt{2})\\&= \boxed{\text{(A) }2}.</cmath>
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To make this problem easier to solve, lets get the radicals out of the denominator. For <math>\frac{1}{2 + \sqrt2}</math>, we will multiply the numerator and denominator by <math>2- \sqrt2</math> so,
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<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, we will  multiply by the conjugate again, <math>\sqrt2 + 2</math>. So that simplifies to
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<math>\frac{1}{\sqrt2 - 2} \cdot \frac{\sqrt2 + 2}{\sqrt2 + 2} \Rightarrow \frac{\sqrt2 + 2}{-2}</math>.
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So now our simplified equation is
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<math>2 + \sqrt2 + \frac{2- \sqrt2}{2} + \frac{\sqrt2 + 2}{-2} \Rightarrow 2+ \sqrt2 + \frac{2 - \sqrt2}{2} - \frac{\sqrt2 +2}{2}</math>
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Bringing everything to the same denominator and combining like terms, we get
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<math>\frac{4 + 2\sqrt2 + 2 - \sqrt2 - \sqrt2 - 2}{2} \Rightarrow \frac{4}{2} \Rightarrow 2 \Rightarrow \boxed{A}</math>
  
 
==See also==
 
==See also==
  
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{{AHSME box|year=1958|num-b=4|num-a=6}}
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{{AHSME 50p box|year=1958|num-b=4|num-a=6}}
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{{MAA Notice}}

Latest revision as of 00:23, 12 March 2017

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

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

See also

1958 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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