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Inequality Question

Discussion in 'DAT Discussions' started by ephesus32, Dec 29, 2008.

  1. ephesus32

    7+ Year Member

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    Can somebody explain this inequality problem: (algebraically)

    What are all the values of x for which (x-2)(x+5) < 0?

    a) 2< x < 5
    b) -2 < x < 5
    c) -5 < x < 2
    d) x < -5
    e) x > 2




    The answer is C.
     
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  3. Sublimation

    5+ Year Member

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    Treat it like a Quadratic form. its like this (x-2)<0 and (X+5)>0 Now solve for X in both. and oyu will end up with C.
     
  4. ephesus32

    7+ Year Member

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    Yeah, I saw that but why do you switch the inequality sign for (x+5)>0 ??
     
  5. FutureInternist

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    So the first step is obviously to solve for x which is pretty easy giving you x = -5 & 2 (which are your critical points). Now you have to choose some test points to see exactly what your solution set is.
    So pick a point less than the least value you have (say -6), pick a point b/w your two critical points (0 is easiest) & then one bigger than your largest value (say 3)
    Now plug each of the test points into the original eqn to see if the solution holds true.
    So for -6 we get -8*-1 which is > 0 so all values less than -5 are out.
    Plugging in 0, you get -2*5 = -10 (which is <0) hence all values b/w -5 & 2 are part of the answer
    Lastly, plugging in 3 --> 1*8=8 (not < 0) so all values > 2 are out.

    So final answer becomes -5<x<2

    Note - There is no need to do the actual multiplication on problems like these....for example for test point -6, just realize that (-6-2) is a (-) & so is (-6+5) and multiplying two negatives gives a positive.
     

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