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Bunuel
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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Lets just simplify this ->

√12 + √27 = 2√3 + 3√3 = 5√3


SMASH that 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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Bunuel
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) \(7\sqrt{3}\)

Simplifying, we have:

√12 + √27

√4 x √3 + √9 x √3 = 2√3 + 3√3 = 5√3

Answer: D
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Bunuel
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) \(7\sqrt{3}\)


=2\sqrt{3}+3\sqrt{3}
=5\sqrt{3}
Answer D
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Bunuel
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) \(7\sqrt{3}\)


\(\sqrt{12}+ \sqrt{27}\)

\(\sqrt{2*2*3}\) + \(\sqrt{3*3*3}\)

2\(\sqrt{3}\) + 3\(\sqrt{3}\)

5\(\sqrt{3}\)

(D)
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Since the basis are not the same, you cannot simplify at the beginning.

\sqrt{12} = \sqrt{2 2 3}, remove the squares = 2 \sqrt{3}

\sqrt{27} = \sqrt{3 3 3}, removed the squares = 3 \sqrt{3}

2\sqrt{3} + 3 \sqrt{3} = 5\sqrt{3}
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