First, x != 0 since we need the denominator to be non-zero. Assuming that x > 0, we can multiply by x without switching the inequality:
x < x^2 - 6 < 5x => x < x^2 - 6 and x^2 - 6 < 5x.
The first one can be rewritten as so, by subtracting x:
0 < x^2 - x - 6 => 0 < (x-3)(x+2) => x > 3 or x<-2.
However, we took x > 0, so only x > 3 is accepted.
The second inequality, is written as:
x^2 - 5x - 6 < 0 => (x-6)(x+1) < 0 => -1 < x < 6
Once again, since x >0, we accept 0< x < 6.
If x > 3 and 0 < x < 6, then 3 < x < 6.
Now, let x < 0. Then x > x^2 - 6 > 5x, since we multiplied by a negative number (x), so the inequality is now the other way around. We can solve them the same way, remembering now that the > changes into < and the < into a >. We arrive at the results:
0 > x^2 - x - 6 => 0 > (x-3)(x+2) => -2 < x < 3
This time, x < 0 so we accept -2 < x < 0.
The second inequality is:
x^2 - 5x - 6 > 0 => (x-6)(x+1) > 0 => x > 6 or x < -1
Since x < 0, we accept x < -1.
Now, we can combine the inequalities:
-2 < x < 3 and x < -1 means -2 < x < -1
So in the end:
3 < x < 6 or -2 < x < -1.
Or in interval notation:
x in (-2, -1)uu(3,6)
Note that you don't have to factor the quadratic polynomials there to solve the quadratic inequality. Simply use the quadratic formula to find the roots, and then if r_1,r_2 are these roots, (r_1 < r_2) then:
The polynomial is positive, if x > r_2 or x < r_1
The polynomial is negative, if r_1 < x < r_2
If the coefficient of x^2 is negative, the above are reversed
And finally, the quadratic formula:
r_(1,2) = (-b +- sqrt(b^2 - 4ac))/(2a)
where a,b,c the coefficients of the polynomial from highest to lowest degree.
