# Solving Rational Inequalities

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• August 11th 2011, 05:43 PM
Dragon08
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!
• August 11th 2011, 05:57 PM
e^(i*pi)
Re: Solving Rational Inequalities
What is the question and what have you tried?

$\dfrac{2x-10}{x} > x+5$ or is it $2x - \dfrac{10}{x} > x+5$
• August 11th 2011, 07:24 PM
SammyS
Re: Solving Rational Inequalities
Quote:

Originally Posted by Dragon08
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: $\dfrac{-2x-10}{x} > x+5\,.$
• August 11th 2011, 07:42 PM
Dragon08
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.
• August 11th 2011, 07:43 PM
Dragon08
Re: Solving Rational Inequalities
To SammyS: Oh, then the the book's answer may be wrong.
• August 12th 2011, 01:19 AM
Siron
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.
• August 12th 2011, 03:57 AM
e^(i*pi)
Re: Solving Rational Inequalities
Quote:

Originally Posted by Siron
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.

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