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

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We need to find x for (x^2-1)/(x^2-4) > -1

Now let us first multiply both the sides by (x^2 -4 ) . Here we assume x^2 -4 is positive.

=> (x^2-1)/(x^2-4) > -1

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

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

=> 2x^2 > 5

=> x^2 > 5/2

=> x > sqrt (5/2) or x < -sqrt(5/2)

For our assumption that x^2 - 4 >0

=> x^2 > 4

=> x > 2 or x < -2

If we assume that x^2 -4 < 0

We get the original inequation as x^2 - 1 < -1* ( x^2 - 4)

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

=> 2x^2 < 5

=> x < sqrt (5/2) or x > - sqrt(5/2)

Also for x^2 -4 < 0

=> x^2 < 4

=> x < 2 or x> -2

Therefore all values of x less than -2, greater than -sqrt(5/2) but less than sqrt(5/2) and those greater than 2 satisfy the inequation.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find x such that (x^2-1)/(x^2-4) > -1

We add 1 to both sides:

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

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

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

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

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

(x-sqrt5/2)(x^2+5/2)/(x-2)(x+2) > 0.

Let f(x) = {(x+sqrt5/2)(x-sqrt5/2)/(x+2)(x-2) > 0.

sqrt(2.5) is around 1.5811etc.

We know -2 < -sqrt(2.5) <   sqr2.5) < 2.

Therefore,

For x < -2 ,  f(x)  = (-)(-)/(-)(-) = positive.

For  -2 < -sqrt2.5 , f(x)  (-)(-)/(+)(-) = -ve.

For  -sqrt2.5 < x < sqrt2.5) , f(x) = (+)(-)/(+)(-) = +ve.

For  sqrt2.5) <x < 2. f(x) = (+)(+)/((+)(-) = -ve.

For x > 2 = f(x) = (+)(+)/(+)(+) = +ve.

Therefore When x <  2 or  -sqrt2.5 < x < sqrt2.5 Or x> 2, the f(x) positive or (x^2-1)/(x^2-4) > -1.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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)