# xWhat is x for (x^2-1)/(x^2-4)>-1?

### 2 Answers | Add Yours

We have to solve (x^2-1)/(x^2-4)>-1

(x^2-1)/(x^2-4)>-1

=> (x^2 - 1) > -1*(x^2 - 4)

=> x^2 - 1 > 4 - x^2

=> 2x^2 - 5 > 0

=> x^2 - 5/2 > 0

=> x^2 > 5/2

**This gives x > sqrt (5/2) and x < -sqrt(5/2)**

We'll put the inequality in the form E(x)>0. For this reason, we'll add 1 both sides, and we'll amplify it with the denominator (x^2-4).

( x^2 - 1 ) / ( x^2 - 4) +1>0

(x^2 - 1 + x^2 - 4)/(x^2-4)>0

We'll combine like terms:

(2*x^2-5)/(x^2-4)>0

We'll note the numerator and denominator as 2 functions:

The numerator: f1(x)=2*x^2-5

The denominator f2(x)=x^2-4

We'll check the monotony of the numerator. In order to do so, first we'll find out the roots of the equation f1(x)=0

2*x^2-5=0 => 2*x^2=5 => x^2=5/2

x1= sqrt (5/2) and x2=-sqrt (5/2)

For f1(x), with a=2>0, between it's roots, f1(x) will be negative and outside the roots, f1(x) will be positive.

f1(x)>0 for x belongs to (-inf,-sqrt(5/2))U(sqrt (5/2),inf)

f2(x)<0 for x belongs to (-sqrt(5/2),sqrt(5/2))

We'll discuss the monotony of the denominator f2(x)=x^2-4

f2(x)= (x-2)(x+2)

(x-2)(x+2)=0

x1=2 and x2=-2

For f2(x), with a=1>0, between it's roots,we'll have the opposed sign to "a" sign, f2(x) will be negative and outside the roots, f2(x) will be positive.

f2(x)>0 for x belongs to (-inf,-2)U(2,inf)

f2(x)<0 for x belongs to (-2,2)

Now, the intervals for E(x)>0 are:

(-inf,-2) U (-sqrt(5/2),sqrt(5/2)) U (2,inf)