Difference between revisions of "1976 AHSME Problems/Problem 27"

(Created page with "== Problem 27 == If <math>N=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}</math>, then <math>N</math> equals <math>\textbf{(A) }1\qquad \t...")
 
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== Problem 27 ==
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==Solution==
  
If <math>N=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}</math>, then <math>N</math> equals
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We will split this problem into two parts: The fraction on the left and the square root on the right.
  
<math>\textbf{(A) }1\qquad
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Starting with the fraction on the left, begin by squaring the numerator and putting a square root around it. It becomes
\textbf{(B) }2\sqrt{2}-1\qquad
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\textbf{(C) }\frac{\sqrt{5}}{2}\qquad
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<math>\frac{\sqrt{(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2})^2}}{\sqrt{\sqrt{5}+1}} = \frac{\sqrt{(\sqrt{5}+2)+(\sqrt{5}-2)+2(
\textbf{(D) }\sqrt{\frac{5}{2}}\qquad
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\sqrt{(\sqrt{5}+2)(\sqrt{5}-2)}}}{\sqrt{\sqrt{5}+1}} = \frac{\sqrt{2\sqrt5+2\sqrt{(\sqrt5)^2-2^2}}}{\sqrt{\sqrt{5}+1}} =
\textbf{(E) }\text{none of these}   </math>
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\frac{\sqrt{2\sqrt5+2\sqrt{1}}}{\sqrt{\sqrt5+1}} =
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\frac{\sqrt{2\sqrt5+2}}{\sqrt{\sqrt5+1}} = \frac{(\sqrt{2})(\sqrt{\sqrt5+1})}{\sqrt{\sqrt5+1}} = \sqrt2</math>.
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Now for the right side.
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<math>\sqrt{3-2\sqrt2} = \sqrt{(1-\sqrt2)^2} = 1-\sqrt2</math>
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Putting it all together gives:
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<math>(\sqrt2)+(1-\sqrt2)=\boxed{(A)  1}.</math>
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~Someonenumber011

Latest revision as of 18:57, 20 March 2020

Solution

We will split this problem into two parts: The fraction on the left and the square root on the right.

Starting with the fraction on the left, begin by squaring the numerator and putting a square root around it. It becomes

$\frac{\sqrt{(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2})^2}}{\sqrt{\sqrt{5}+1}} = \frac{\sqrt{(\sqrt{5}+2)+(\sqrt{5}-2)+2( \sqrt{(\sqrt{5}+2)(\sqrt{5}-2)}}}{\sqrt{\sqrt{5}+1}} = \frac{\sqrt{2\sqrt5+2\sqrt{(\sqrt5)^2-2^2}}}{\sqrt{\sqrt{5}+1}} =  \frac{\sqrt{2\sqrt5+2\sqrt{1}}}{\sqrt{\sqrt5+1}} = \frac{\sqrt{2\sqrt5+2}}{\sqrt{\sqrt5+1}} = \frac{(\sqrt{2})(\sqrt{\sqrt5+1})}{\sqrt{\sqrt5+1}} = \sqrt2$.

Now for the right side.

$\sqrt{3-2\sqrt2} = \sqrt{(1-\sqrt2)^2} = 1-\sqrt2$

Putting it all together gives:

$(\sqrt2)+(1-\sqrt2)=\boxed{(A)  1}.$

~Someonenumber011

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