Bunuel wrote:

What is the sum of \(\sqrt{12}+ \sqrt{27}\) ?

(A) \(\sqrt{29}\)

(B) \(3\sqrt{5}\)

(C) \(13\sqrt{3}\)

(D) \(5\sqrt{3}\)

(E) \(7\sqrt{3}\)

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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