ABC26GN1393 · Square Roots and Cube Roots
Subject: General Aptitude · Chapter: Square Roots and Cube Roots · Exam: · Marks: · Difficulty:
By how much does $\sqrt{12}+\sqrt{18}$ exceed $\sqrt{3}+\sqrt{2}$ ?
(a)$\sqrt{2}-4 \sqrt{3}$
(b)$\sqrt{3}+2 \sqrt{2}$
(c)$2(\sqrt{3}-\sqrt{2})$
(d)$3(\sqrt{3}-\sqrt{2})$
Answer
Explanation
$$\begin{aligned} (\sqrt{12}+\sqrt{18})- & (\sqrt{3}+\sqrt{2})=(\sqrt{4 \times 3}+\sqrt{9 \times 2})-(\sqrt{3}+\sqrt{2}) \\ = & (2 \sqrt{3}+3 \sqrt{2})-(\sqrt{3}+\sqrt{2}) \\ = & (2 \sqrt{3}-\sqrt{3})+(3 \sqrt{2}-\sqrt{2})=\sqrt{3}+2 \sqrt{2} . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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