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GRE math question

  1. Sep 3, 2015 #1
    I was going through some GRE Quantitative Practice Tests, and found this question. Maybe I'm not understanding the question right (not a native speaker, and have never solved math problems in English), but choice B doesn't seem correct to me.
    Solutions for x (Quantity A) are 6 and -1, and for y(Quantity B) are -3 and -3. Meaning, -3<6 and -3<-1.
    Shouldn't A be correct?
    Also, there's an explanation at the bottom which is correct in general but can't seem to understand how it proves the correct choice.

    X4x7Qis.png
     
  2. jcsd
  3. Sep 3, 2015 #2

    jedishrfu

    Staff: Mentor

    Yeah, I think you're right. The answer makes no sense. Also the explanation seems to be geared toward a problem of quantity A=c value and quantity B=k value.

    Perhaps you can send it to someone at the GRE center for clarification.
     
  4. Sep 3, 2015 #3

    Mark44

    Staff: Mentor

    The answer given doesn't match the question, as far as I can see, so my guess is that there is a significant typo in the question. In the answer shown, they talk about factors (x + 2) and (x + 3).
    The equation in x that they give is equivalent to x2 - 5x -6 = 0, or (x - 6)(x + 1) = 0, so x = 6 or x = -1.
    The equation in y is equivalent to (y + 3)2 = 0, so y = -3.

    For the two equations, both values of x are larger than the single y value.
     
    Last edited: Sep 3, 2015
  5. Sep 3, 2015 #4
    Unless I miscalculated, if ## y = -3 ## then ## x^2 - 5x + 18 = 0 ## which gives ## x ## the value ## {5 \pm i\sqrt{47}} \over {2} ##. Since reals and complexes cannot be compared, I would say D! (I think there's a mistake in the printing.)
     
  6. Sep 3, 2015 #5

    Mark44

    Staff: Mentor

    I don't see how you got your last equation. The two equations are independent of one another, so how does a value for y result in ##x^2 - 5x + 18 = 0##?
     
  7. Sep 3, 2015 #6
    <EDIT> Ignore this post. Misread the scanned problem. </EDIT>

    Since you have ## E_1 = E_2 = E_3 ##, you merely solve ## E_2 = E_3 ## first, and then substitute your value of ## y ## back into ## E_1 = E_2 ##.

    ## E_2 = E_3 : -y^2 + 3y = 9y + 9 ##
    ## -y^2 + 3y = 9y + 9 \rightarrow (y + 3)^2 = 0 \rightarrow y = -3 ##

    ## E_1 = E_2 : x^2 - 5x = -y^2 + 3y ##
    ## x^2 - 5x = -y^2 + 3y \rightarrow x^2 - 5x = -1 \cdot (-3)^2 + 3 \cdot (-3) \rightarrow x^2 - 5x = -18 \rightarrow x^2 - 5x + 18 = 0 ##
     
    Last edited: Sep 3, 2015
  8. Sep 3, 2015 #7
    Oh, wait, one could take the magnitude of both numbers!
     
  9. Sep 3, 2015 #8

    Mark44

    Staff: Mentor

    I don't think the intent was to find a simultaneous solution; i.e., a solution (x, y). My take is that the intent was merely to compare the solutions to the two equations, in which case answer A would be the correct response.
     
  10. Sep 3, 2015 #9
    Sorry. Small screen display. I saw the problem wrong. :D
     
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