8. Exponents and Roots of Numbers

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Another apporach to solve this problem is as follows: If x = \(\sqrt{12}+ \sqrt{27}\)

Squaring, we get \(x^2 = (\sqrt{12}+ \sqrt{27})^2 = 12 + 27 + 2\sqrt{12}\sqrt{27} = 39 + 2\sqrt{4*81} = 39 + 2*2*9 = 39 + 36 = 75\)

Therefore, taking square root, we will get x = \(\sqrt{3*5*5} = 5\sqrt{3}\)(Option D)
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\(\sqrt{12}\) = 2 \(\sqrt{3}\)
\(\sqrt{27}\)= 3 \(\sqrt{3}\)

\(\sqrt{12}\) + \(\sqrt{27}\) =\(2\sqrt{3}\) + \(3\sqrt{3}\) = 5 \(\sqrt{3}\)

Answer: Option D
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=2\sqrt{3}+3\sqrt{3}
=5\sqrt{3}
Answer D