ABC26GN1959 · Surds and Indices
Subject: General Aptitude · Chapter: Surds and Indices · Exam: 2010 · Marks: · Difficulty:
$\sqrt{6-4 \sqrt{3}+\sqrt{16-8 \sqrt{3}}}$ is equal to
(a)$1-\sqrt{3}$
(b)$\sqrt{3}-1$
(c)$2(2-\sqrt{3})$
(d)$2(2+\sqrt{3})$
Answer
Explanation
$$\begin{aligned} & \sqrt{6-4 \sqrt{3}}+\sqrt{16-8 \sqrt{3}}=\sqrt{6-4 \sqrt{3}+\sqrt{12+4-8 \sqrt{3}}} \\ &= \sqrt{6-4 \sqrt{3}+\sqrt{(2 \sqrt{3})^{2}+(2)^{2}-2 \times 2 \sqrt{3} \times 2}} \\ &= \sqrt{6-4 \sqrt{3}+\sqrt{(2 \sqrt{3}-2)^{2}}}=\sqrt{6-4 \sqrt{3}+2 \sqrt{3}-2} \\ &= \sqrt{(\sqrt{3})^{2}+(1)^{2}-2 \times \sqrt{3} \times 1}=\sqrt{(\sqrt{3}-1)^{2}}=\sqrt{3}-1 \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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