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

m (Solution: (fixed formatting))
(Centered the important equation: https://files.eric.ed.gov/fulltext/ED239856.pdf Also, I am thinking of making the solution clearer using algebra.)
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==Problem==
 
==Problem==
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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If <cmath>N=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}},</cmath> then <math>N</math> equals
 +
 
 +
<math>\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}  </math>
  
<math>\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}</math>
 
 
==Solution==
 
==Solution==
  
We will split this problem into two parts: The fraction on the left and the square root on the right.
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~Someonenumber011 (Solution)
 
 
Starting with the fraction on the left, begin by squaring the numerator and putting a square root around it. It becomes
 
 
 
<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}}</math>
 
 
 
 
 
<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>.
 
 
 
 
 
Now for the right side.
 
 
 
<math>\sqrt{3-2\sqrt2} = \sqrt{(1-\sqrt2)^2} = 1-\sqrt2</math>
 
 
 
Putting it all together gives:
 
 
 
<math>(\sqrt2)+(1-\sqrt2)=\boxed{(A)  1}.</math>
 
  
~Someonenumber011
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~MRENTHUSIASM (Revision)
  
 
== See also ==
 
== See also ==

Revision as of 23:49, 7 September 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

~Someonenumber011 (Solution)

~MRENTHUSIASM (Revision)

See also

1976 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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