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  # If 2x/ (root(-3x^2 + 27)) is not a real number
  
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  If 2x/ (root(-3x^2 + 27)) is not a real number  [#permalink]
  
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  10 Jul 2018, 01:48
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  Question Stats:
  
  53% (01:16) correct 47% (01:36) wrong based on 133 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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  If 2x/ (root(-3x^2 + 27)) is not a real number  [#permalink]
  
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  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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  Re: If 2x/ (root(-3x^2 + 27)) is not a real number  [#permalink]
  
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  10 Jul 2018, 04:54
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  Karthik200 wrote:
  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. $$-3 \leq x \leq 3$$
  
  Since the GMAT is constrained to real numbers, I would disregard this problem.
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  Re: If 2x/ (root(-3x^2 + 27)) is not a real number  [#permalink]
  
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  10 Jul 2018, 05:03
  GMATGuruNY wrote:
  
  Since the GMAT is constrained to real numbers, I would disregard this problem.
  
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  Re: If 2x/ (root(-3x^2 + 27)) is not a real number  [#permalink]
  
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  15 Jul 2018, 10:13
  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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  Re: If 2x/ (root(-3x^2 + 27)) is not a real number  [#permalink]
  
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  18 Jul 2018, 18:22
  workout wrote:
  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
  
  Thanks workout for the nice explanation
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  Re: If 2x/ (root(-3x^2 + 27)) is not a real number  [#permalink]
  
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  18 Jul 2018, 19:56
  -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
  
  Grade school algebra questions rly comfortable easy
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  Re: If 2x/ (root(-3x^2 + 27)) is not a real number  [#permalink]
  
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  21 Jan 2020, 09:20
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  Re: If 2x/ (root(-3x^2 + 27)) is not a real number   [#permalink] 21 Jan 2020, 09:20
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