sqrt 2 and 3

Algebra Level 2

What is x x ? 2 + 3 × 2 + 2 + 3 × 2 + 2 + 2 + 3 × 2 2 + 2 + 3 = x \sqrt{2+\sqrt{3}}\times \sqrt{2+\sqrt{2+\sqrt{3}}}\times \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\times \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}=x


The answer is 1.000.

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2 solutions

Chew-Seong Cheong
Feb 27, 2018

x = 2 + 3 × 2 + 2 + 3 × 2 + 2 + 2 + 3 × 2 2 + 2 + 3 = 2 + 3 × 2 + 2 + 3 × 2 2 ( 2 + 2 + 3 ) 2 = 2 + 3 × 2 + 2 + 3 × 2 2 + 3 = 2 + 3 × 2 2 ( 2 + 3 ) 2 = 2 + 3 × 2 3 = 2 2 ( 3 ) 2 = 4 3 = 1 \begin{aligned} x & = \sqrt{2+\sqrt 3} \times \sqrt{2+\sqrt{2+\sqrt 3}} \times \sqrt{2+\sqrt{2+\sqrt{2+\sqrt 3}}} \times \sqrt{2-\sqrt{2+\sqrt{2+\sqrt 3}}} \\ & = \sqrt{2+\sqrt 3} \times \sqrt{2+\sqrt{2+\sqrt 3}} \times \sqrt{2^2-\left(\sqrt{2+\sqrt{2+\sqrt 3}}\right)^2} \\ & = \sqrt{2+\sqrt 3} \times \sqrt{2+\sqrt{2+\sqrt 3}} \times \sqrt{2-\sqrt{2+\sqrt 3}} \\ & = \sqrt{2+\sqrt 3} \times \sqrt{2^2-\left(\sqrt{2+\sqrt 3}\right)^2} \\ & = \sqrt{2+\sqrt 3} \times \sqrt{2-\sqrt 3} \\ & = \sqrt{2^2-\left(\sqrt 3\right)^2} \\ & = \sqrt{4-3} \\ & = \boxed{1} \end{aligned}

Naren Bhandari
Feb 27, 2018

Recall that tan 75 ° = 2 + 3 \begin{aligned}\tan75° =2 +\sqrt 3\end{aligned} and tan 1 5 = 2 3 \tan15^{\circ} = 2-\sqrt3 . Rewriting the given expression as x = tan 75 ° 2 + tan 75 ° 2 + 2 + tan 75 ° 2 2 + tan 75 ° x = 2 tan 75 ° + tan 75 ° 2 2 ( 2 + tan 75 ° ) 2 x = 2 tan 75 ° + tan 2 75 ° 2 tan 75 ° x = tan 75 ° tan 15 ° x = tan 75 ° cot 75 ° = 1 \begin{aligned} & x = \sqrt{\tan75°} \sqrt{2 + \sqrt{\tan75°}}\sqrt{2 + \sqrt{2+\sqrt{\tan75°}}}\sqrt{2-\sqrt{2+\sqrt{\tan75°}}} \\& x = \sqrt{2\tan{75°} + \tan 75°} \sqrt{2^2 - (\sqrt{2 + \sqrt{\tan75°}}})^2 \\& x =\sqrt{2\tan75° +\tan^2 75°} \sqrt{2 - \sqrt{\tan 75°}} \\& x = \sqrt{\tan75°} \sqrt{\tan{15°}} \\& x = \sqrt{\tan75°}\sqrt{\cot75°} =1 \end{aligned}

Thanks for sharing it. But you can see that you don't have to use neither tan \tan nor cot \cot .

Áron Bán-Szabó - 3 years, 3 months ago

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Your welcome !! :) Yeah I see the another solution as well. I like your problems much . :)

Naren Bhandari - 3 years, 3 months ago

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