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Difference between revisions of "1976 AHSME Problems/Problem 27"
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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(
 
<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(
−
\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}} =  
+
\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}}</math>
−
\frac{\sqrt{2\sqrt5+2\sqrt{1}}}{\sqrt{\sqrt5+1}} =
+
 
 +
 
 +
<math>=\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</math>.
 
\frac{\sqrt{2\sqrt5+2}}{\sqrt{\sqrt5+1}} = \frac{(\sqrt{2})(\sqrt{\sqrt5+1})}{\sqrt{\sqrt5+1}} = \sqrt2</math>.
 +
  
 
Now for the right side.
 
Now for the right side.
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~Someonenumber011
 
~Someonenumber011
 +
 
== See also ==
 
== See also ==
  

Revision as of 19:59, 19 July 2021

Problem

If $N=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}$, then $N$ equals

$\textbf{(A) }1\qquad \textbf{(B) }2\sqrt{2}-1\qquad \textbf{(C) }\frac{\sqrt{5}}{2}\qquad \textbf{(D) }\sqrt{\frac{5}{2}}\qquad \textbf{(E) }\text{none of these}$

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

See also

1976 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
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
All AHSME Problems and Solutions

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