How do I solve the inequality $x<x^2-12<4x$?

Question

So first I considered $x < x^2 -12$ so I get $0 < x^2 - x -12$ which is $0<(x+3)(x-4)$

after this I don't know where to go

Again, I considered $x^2 - 12< 4x$ which is $x^2 - 4x - 12<0$ so $(x+2)(x-6)<0$

Again same issue, I don't know where to after that.

Answer 1

You have to solve the system of quadratic inequalities \begin{cases} x^2-x-12>0 ,\\ x^2-4x-12<0 . \end{cases} The first quadratic polynomial has roots $4$ and $-3$, hence the solutions of the first inequation is $$S_1=(-\infty,-3)\cup(4,\infty).$$ As to the second quadratic polynomial, it has roots $6$ and $-2$, so the solution of the second inequation is $$S_2=(-2, 6),$$ and the solutions of the system is $$S_1\cap S_2=(4,6).$$

Answer 2

$$0 < (x+3)(x-4)$$

is equivalent to $x < -3$ or $x >4$. To see this sketch the convex quadratic curve and see where is it positive.

Similarly, $$(x+2)(x-6) < 0$$ is equivalent to $$-2 < x<6.$$

Intersecting them, the region of interest is

$$4 < x< 6.$$