1. 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 can some one please show me how this answer is arrived at..

2. Originally Posted by sigma1
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 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.

3. 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.