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

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<math>(\sqrt2)+(1-\sqrt2)=\boxed{(A)  1}.</math>
 
<math>(\sqrt2)+(1-\sqrt2)=\boxed{(A)  1}.</math>
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~Someonenumber011

Revision as of 17:57, 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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