Find the range of real numbers x satisfying both inequalities at once: x^2 + x - 6 < 0 and x^2 + 4x - 12 > 0.

Step-by-step solution

1. 1

   Solve each inequality on its own before combining. A system of inequalities is an intersection, so the plan is: get the solution set of each line, then keep only the $x$ values in both. Start by factoring the first:

   $x^{2}+x-6=(x+3)(x-2)$

   The roots are $x=-3$ and $x=2$.

2. 2

   Read off the first solution set from the parabola's shape. The leading coefficient is $+1$, so the parabola opens upward and dips below the axis strictly between its roots:

   $(x+3)(x-2)<0\quad\Longleftrightarrow\quad -3<x<2$

3. 3

   Factor and solve the second inequality. Look for the pair multiplying to $-12$ and adding to $4$, which is $6$ and $-2$:

   $x^{2}+4x-12=(x+6)(x-2)$

   The roots are $x=-6$ and $x=2$. This time we want the expression above the axis, which for an upward parabola happens outside the roots:

   $x<-6\quad\text{or}\quad x>2$

4. 4

   Intersect the two sets. The candidate values must lie in $(-3,2)$ and also in $(-\infty,-6)\cup(2,\infty)$:

   $(-3,2)\cap(-\infty,-6)=\varnothing$

   because every number below $-6$ is already below $-3$. And

   $(-3,2)\cap(2,\infty)=\varnothing$

   because $(-3,2)$ stops just short of $2$ while the other set starts just past it. The shared root $x=2$ is excluded by both strict inequalities, so it cannot rescue the intersection.

5. 5

   State the conclusion. No real number satisfies both inequalities, so the solution set is empty:

   $x\in\varnothing$

   A scan of every $x$ from $-10$ to $10$ in steps of $0.001$ turns up no value passing both tests, which matches this result.