Difference between revisions of "1958 AHSME Problems/Problem 5"
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==Solution== | ==Solution== | ||
− | {{ | + | 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, |
+ | |||
+ | <math>\frac{1}{2 + \sqrt2} \cdot \frac{2 - \sqrt2}{2 - \sqrt2} \Rightarrow \frac{2 - \sqrt2}{2}</math>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | So now our simplified equation is | ||
+ | |||
+ | <math>2 + \sqrt2 + \frac{2- \sqrt2}{2} + \frac{\sqrt2 + 2}{-2} \Rightarrow 2+ \sqrt2 + \frac{2 - \sqrt2}{2} - \frac{\sqrt2 +2}{2}</math> | ||
+ | |||
+ | Bringing everything to the same denominator and combining like terms, we get | ||
+ | |||
+ | <math>\frac{4 + 2\sqrt2 + 2 - \sqrt2 - \sqrt2 - 2}{2} \Rightarrow \frac{4}{2} \Rightarrow 2 \Rightarrow \boxed{A}</math> | ||
==See also== | ==See also== | ||
− | {{AHSME box|year=1958|num-b=4|num-a=6}} | + | {{AHSME 50p box|year=1958|num-b=4|num-a=6}} |
+ | {{MAA Notice}} |
Latest revision as of 00:23, 12 March 2017
Problem
The expression equals:
Solution
To make this problem easier to solve, lets get the radicals out of the denominator. For , we will multiply the numerator and denominator by so,
.
Now, the other fraction we need to get the radical out of the denominator is . Here, we will multiply by the conjugate again, . So that simplifies to
.
So now our simplified equation is
Bringing everything to the same denominator and combining like terms, we get
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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