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Question: Answered & Verified by Expert
$\sqrt{2+\sqrt{5}-\sqrt{6-3 \sqrt{5}+\sqrt{14-6 \sqrt{5}}}}$ is equal to
MathematicsBasic of MathematicsTS EAMCETTS EAMCET 2007
Options:
  • A $1$
  • B $2$
  • C $3$
  • D $4$
Solution:
2906 Upvotes Verified Answer
The correct answer is: $2$
Now,
$\begin{aligned} \sqrt{2+} & \sqrt{5}-\sqrt{6-3 \sqrt{5}+\sqrt{14-6 \sqrt{5}}} \\ & =\sqrt{2+\sqrt{5}-\sqrt{6-3 \sqrt{5}+\sqrt{(9+5-6 \sqrt{5})}}} \\ & =\sqrt{2+\sqrt{5}-\sqrt{6-3 \sqrt{5}+\sqrt{(3-\sqrt{5})^2}}} \\ & =\sqrt{2+\sqrt{5}-\sqrt{9-4 \sqrt{5}}} \\ & =\sqrt{2+\sqrt{5}-\sqrt{(-2+\sqrt{5})^2}} \\ & =\sqrt{2+\sqrt{5}+2-\sqrt{5}} \\ & =2\end{aligned}$

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