# inequality question

• Jun 6th 2010, 07:51 PM
sigma1
inequality question
hello i was asked to solve the real value of x for which $\displaystyle \frac{x+1}{x-2}< 0$

i got$\displaystyle x< 2$ and $\displaystyle x < -1$ but the answer in the book is $\displaystyle -1<x<2$ can some one please show me how this answer is arrived at..
• Jun 6th 2010, 07:59 PM
TheEmptySet
Quote:

Originally Posted by sigma1
hello i was asked to solve the real value of x for which $\displaystyle \frac{x+1}{x-2}< 0$

i got$\displaystyle x< 2$ and $\displaystyle x < -1$ but the answer in the book is $\displaystyle -1<x<2$ 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 $\displaystyle x=-1$ and $\displaystyle x=2$

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

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

Now just check test points in each regions

So if we check $\displaystyle x=-2$ we get

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

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.
• Jun 7th 2010, 06:23 AM
Krizalid
as said above we have $\displaystyle x\ne2,$ now multiply both sides by $\displaystyle (x-2)^2$ and the inequality becomes $\displaystyle (x+1)(x-2)<0\implies x^2-x-2<0\implies 4x^2-4x-8<0\implies(2x-1)^2<9$ thus $\displaystyle |2x-1|<3$ which yields $\displaystyle -3<2x-1<3$ and we're done.