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Math Help - inequality question

  1. #1
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    inequality question

    hello i was asked to solve the real value of x for which \frac{x+1}{x-2}< 0

    i got  x< 2 and x < -1 but the answer in the book is -1<x<2 can some one please show me how this answer is arrived at..
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by sigma1 View Post
    hello i was asked to solve the real value of x for which \frac{x+1}{x-2}< 0

    i got  x< 2 and x < -1 but the answer in the book is -1<x<2 can some one please show me how this answer is arrived at..

    So first you need to solve for all the zero points of the numerator and denominator

    They are x=-1 and x=2

    This divides the number line into three disjoint parts i.e

    x \le -1 and -1 \le x < 2 and x > 2

    Now just check test points in each regions

    So if we check x=-2 we get

    \frac{-2+1}{-2-2}=\frac{1}{4} \not < 0

    So the first set is not a solution. Now just check the other two and you will find that the middle one is the only one that checks.
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  3. #3
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    as said above we have x\ne2, now multiply both sides by (x-2)^2 and the inequality becomes (x+1)(x-2)<0\implies x^2-x-2<0\implies 4x^2-4x-8<0\implies(2x-1)^2<9 thus |2x-1|<3 which yields -3<2x-1<3 and we're done.
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