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hazelnut
vikasp99
\((√11 + √11 + √11)^2\) =

(A) 363
(B) 121
(C) 99
(D) 66
(E) 33

\((√11 + √11 + √11)^2\) = \((3\sqrt{11})^{2}\) = \(3^2*11\)=\(99\)

I am confused. I got 33. What am I missing?

\((√11 + √11 + √11)\) = \(√33\) right? Do the parenthesis first, then the exponent. \(√33^2\)=33
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hazelnut
vikasp99
\((√11 + √11 + √11)^2\) =

(A) 363
(B) 121
(C) 99
(D) 66
(E) 33

\((√11 + √11 + √11)^2\) = \((3\sqrt{11})^{2}\) = \(3^2*11\)=\(99\)

I am confused. I got 33. What am I missing?

\((√11 + √11 + √11)\) = \(√33\) right? Do the parenthesis first, then the exponent. \(√33^2\)=33

Let me ask you what does x + x + x equal to? 3x, right? The same way, \(√11 + √11 + √11=3\sqrt{11}\). You would be right if it were \(\sqrt{11 + 11 +11}=\sqrt{33}\).

Does this make sense?
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Yep, that makes perfect sense. Thanks Bunuel. I see I did not know the \(\sqrt{a}\)+\(\sqrt{b}\) \(\neq\) \(\sqrt{a+b}\) rule from GMATclub book.
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vikasp99
(√11 + √11 + √11)^2 =

(A) 363
(B) 121
(C) 99
(D) 66
(E) 33

(√11 + √11 + √11)^2

(3√11)^2 = 3^2 x 11 = 99

Answer: C
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