ABC26GN1432 · Square Roots and Cube Roots
Subject: General Aptitude · Chapter: Square Roots and Cube Roots · Exam: · Marks: · Difficulty:
$\sqrt{2+\sqrt{3}} \cdot \sqrt{2+\sqrt{2+\sqrt{3}}} \cdot \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}$ $\cdot \sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}$ is equal to
(a)1
(b)2
(c)4
(d)$\sqrt{6}$
Answer
Explanation
Given expression $$\begin{aligned} & =\sqrt{2+\sqrt{3}} \cdot \sqrt{2+\sqrt{2+\sqrt{3}}} \cdot \sqrt{2^{2}-(\sqrt{2+\sqrt{2+\sqrt{3}}})^{2}} \\ & =\sqrt{2+\sqrt{3}} \cdot \sqrt{2+\sqrt{2+\sqrt{3}}} \cdot \sqrt{4-(2+\sqrt{2+\sqrt{3})}} \\ & =\sqrt{2+\sqrt{3}} \cdot \sqrt{2+\sqrt{2+\sqrt{3}}} \cdot \sqrt{2-\sqrt{2+\sqrt{3}}} \\ & =\sqrt{2+\sqrt{3}} \cdot \sqrt{2^{2}-(\sqrt{2+\sqrt{3}})^{2}}=\sqrt{2+\sqrt{3}} \cdot \sqrt{2-\sqrt{3}} \\ & =\sqrt{2^{2}-(\sqrt{3})^{2}}=\sqrt{4-3}=\sqrt{1}=1 . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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