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Math Help - Solving Rational Inequalities

  1. #1
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    Solving Rational Inequalities

    Solve the inequality: 2x-10/x >x+5
    I tried it and got an unfactorable polynomial: x^2+3x+10, but the answer says it's x<-5 and -2<x<0.
    Thanks in advance!
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  2. #2
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    Re: Solving Rational Inequalities

    What is the question and what have you tried?

    \dfrac{2x-10}{x} > x+5 or is it 2x - \dfrac{10}{x} > x+5
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  3. #3
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    Re: Solving Rational Inequalities

    Quote Originally Posted by Dragon08 View Post
    Solve the inequality: 2x-10/x >x+5
    I tried it and got an unfactorable polynomial: x^2+3x+10, but the answer says it's x<-5 and -2<x<0.
    Thanks in advance!
    To get the book answer, the problem would have to be equivalent to solving: \dfrac{-2x-10}{x} > x+5\,.
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    Re: Solving Rational Inequalities

    It's the first one, and I made the moved x+5 onto the left side and made a common denominator. The resulting polynomial numerator is unfactorable.
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  5. #5
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    Re: Solving Rational Inequalities

    To SammyS: Oh, then the the book's answer may be wrong.
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  6. #6
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    Re: Solving Rational Inequalities

    First of all, like SammyS said it has to be -2x in the numerator of the RS, because if it's 2x then you'll get a negative discriminant and so no roots in \mathbb{R}.

    Rational inequality
    \frac{-2x-10}{x}>x+5 and x\neq 0

    Solve:
    \frac{-2x-10}{x}>x+5 \Leftrightarrow -2x-10>x(x+5) \Leftrightarrow -2x-10>x^2+5x \Leftrightarrow -x^2-7x-10>0 \Leftrightarrow -(x+5)(x+2)>0

    Now you can make a sign table to determine the solution.
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    Re: Solving Rational Inequalities

    Quote Originally Posted by Siron View Post
    First of all, like SammyS said it has to be -2x in the numerator of the RS, because if it's 2x then you'll get a negative discriminant and so no roots in \mathbb{R}.

    Rational inequality
    \frac{-2x-10}{x}>x+5 and x\neq 0

    Solve:
    \frac{-2x-10}{x}>x+5 \Leftrightarrow -2x-10>x(x+5) \Leftrightarrow -2x-10>x^2+5x \Leftrightarrow -x^2-7x-10>0 \Leftrightarrow -(x+5)(x+2)>0

    Now you can make a sign table to determine the solution.
    Good spot on the -2x instead of 2x (Clap)

    Alternatively, if x < 0 then we need to change the direction of the inequality when multiplying through by x.

    -2x-10 < x^2+5x \Longleftrightarrow x^2+7x+10 > 0 \Longleftrightarrow (x+2)(x+5) > 0
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