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If 2x/ (root(-3x^2 + 27)) is not a real number 10 Jul 2018, 01:48
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55% (hard)

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57% (01:29) correct 43% (01:40) wrong based on 509 sessions

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If \(\frac{2x}{\sqrt{-3x^2 + 27}}\) is not a real number, which of the following specifies all the possible values of x?

A. \(x\leq-3\) or \(x\geq3\)
B. \(x\leq-3\)
C. \(x\geq3\)
D. \(x\leq-4\) or \(x\geq4\)
E. \(-3 \leq x \leq 3\)

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A
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Nikhil
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If 2x/ (root(-3x^2 + 27)) is not a real number 10 Jul 2018, 01:55
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GIven \(\frac{2x}{\sqrt{-3x^2 + 27}}\) is not a real number

=> The number inside the square root sign is < 0. Also since the square root is in denominator, the number is not a real number if denominator is 0

=> \(-3x^2 + 27 \leq 0\)

=> \(3x^2 - 27 \geq 0\)

=> \(3x^2 \geq 27\)

=> \(x^2 \geq 9\)

=> \(x\leq-3\) or \(x\geq3\)

Hence option A
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If 2x/ (root(-3x^2 + 27)) is not a real number 10 Jul 2018, 04:54
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Karthik200
If \(\frac{2x}{\sqrt{-3x^2 + 27}}\) is not a real number, which of the following specifies all the possible values of x?

A. \(x\leq-3\) or \(x\geq3\)
B. \(x\leq-3\)
C. \(x\geq3\)
D. \(x\leq-4\) or \(x\geq4\)
E. \(-3 \leq x \leq 3\)
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Since the GMAT is constrained to real numbers, I would disregard this problem.
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If 2x/ (root(-3x^2 + 27)) is not a real number 10 Jul 2018, 05:03
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GMATGuruNY


Since the GMAT is constrained to real numbers, I would disregard this problem.

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Good to know about this constraint. Thank you.
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If 2x/ (root(-3x^2 + 27)) is not a real number 15 Jul 2018, 10:13
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Although I got this question right, I am just curious to check is this a GMAT like question? I was told by someone that imaginary numbers are not tasted on GMAT.
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If 2x/ (root(-3x^2 + 27)) is not a real number 18 Jul 2018, 19:56
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-3x^2+27<0
3x^2-27>0
3(x^2-9)>0
x^2-9>0
(x+3)(x-3)>0

x+3>0 and x-3>0
x>-3 and x>3

x+3<0 and x-3<0
x<-3 and x<3
Answer A

Grade school algebra questions rly comfortable easy :cool:
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If 2x/ (root(-3x^2 + 27)) is not a real number 19 Sep 2023, 02:07
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I have a question regarding the answer here :

\(\sqrt{-3x^2 + 27}\) is in the denominator and we know that anything in the denominator cannot be 0 else the result will be undefined.
Now why shouldn't we equate this to 0.
\(\sqrt{-3x^2 + 27}\) = 0
Squaring both sides
we get :
\(-3x^2 + 27 = 0\)
\(-3x^2 = -27\)
\(x^2 = 9\)
x = +3 or -3

Now my question here is that :

From this I understand that when x = 3 or -3 the denominator will be 0 and thus undefined.
I know that for me the condition "the number is not a real number" makes this question slightly tricky.
shouldn't we exclude the exact values of x = 3 or -3 from this range making the answer x>3 U x<-3. and NOT x=>3 U x<=-3.
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If 2x/ (root(-3x^2 + 27)) is not a real number 07 Feb 2025, 05:10
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KarthikMBA2019
If \(\frac{2x}{\sqrt{-3x^2 + 27}}\) is not a real number, which of the following specifies all the possible values of x?

A. \(x\leq-3\) or \(x\geq3\)
B. \(x\leq-3\)
C. \(x\geq3\)
D. \(x\leq-4\) or \(x\geq4\)
E. \(-3 \leq x \leq 3\)
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Correct answer should and must be D. If you have 0 in the denominator it makes the number undefined or infinite but not complex or not real. Only thing which makes a number complex or not real is negative square root which in this case is for x>3 and x<-3

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