ABC26GN1494 · Square Roots and Cube Roots

Subject: General Aptitude · Chapter: Square Roots and Cube Roots · Exam: · Marks: · Difficulty:

$\sqrt{10+2 \sqrt{6}+2 \sqrt{10}+2 \sqrt{15}}$ is equal to
(a)$(\sqrt{2}+\sqrt{3}+\sqrt{5})$
(b)$(\sqrt{2}+\sqrt{3}-\sqrt{5})$
(c)$(\sqrt{2}+\sqrt{5}-\sqrt{3})$
(d)None of these [DMRC-Customer Relationship Assistant (CRA) Exam, 2016]
Answer
Answer (as printed): A
Explanation
Given $\sqrt{10+2 \sqrt{6}+2 \sqrt{10}+2 \sqrt{15}}$ $$\begin{aligned} & =\sqrt{10+2 \times \sqrt{3} \times \sqrt{2}+2 \times \sqrt{2} \times \sqrt{5}+2 \times \sqrt{3} \times \sqrt{5}} \\ & =\sqrt{2+3+5+2 \times \sqrt{2} \times \sqrt{3}+2 \times \sqrt{5} \times \sqrt{2}+2 \times \sqrt{5} \times \sqrt{3}} \\ & =\sqrt{(\sqrt{2})^{2}+(\sqrt{3})^{2}+(\sqrt{5})^{2}+2 \times \sqrt{2} \times \sqrt{3}+2 \times \sqrt{5} \times \sqrt{2}+2 \times \sqrt{5} \times \sqrt{3}} \\ & \quad \quad\left\{\because a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a=(a+b+c)^{2}\right\} \\ & =\sqrt{(\sqrt{2}+\sqrt{3}+\sqrt{5})^{2}} \\ & =(\sqrt{2}+\sqrt{3}+\sqrt{5}) \end{aligned}$$

Explanation as extracted from the printed page; notation may be imperfect.

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