Solve x^2 - 3x - 10 >= 0 and x^2 - 1 <= 0, and find the values of x satisfying both.

Step-by-step solution

1. 1

   Factor the first quadratic. Two numbers with product $-10$ and sum $-3$ are $-5$ and $+2$:

   $x^{2}-3x-10=(x-5)(x+2)\ge 0$

   Critical points: $x=-2$ and $x=5$.

2. 2

   Read the first solution set off the parabola. The leading coefficient is positive, so this upward parabola is at or above the axis outside its roots (test $x=0$: $-10<0$, so the middle interval fails):

   $(-\infty,-2]\cup[5,\infty)$

   Both endpoints are included because the inequality is non-strict.

3. 3

   Factor the second quadratic as a difference of squares.

   $x^{2}-1=(x-1)(x+1)\le 0$

   Critical points: $x=-1$ and $x=1$.

4. 4

   Read the second solution set. An upward parabola is at or below the axis between its roots (test $x=0$: $-1\le 0$ ✓):

   $[-1,1]$

5. 5

   Intersect the two sets. Overlay them on one number line: the second set lives entirely inside $(-2,5)$, which is precisely the interval the first set excludes. Formally $[-1,1]\cap(-\infty,-2]=\varnothing$ and $[-1,1]\cap[5,\infty)=\varnothing$, so

   $\left((-\infty,-2]\cup[5,\infty)\right)\cap[-1,1]=\varnothing$

6. 6

   State both answers. Individually: $(-\infty,-2]\cup[5,\infty)$ and $[-1,1]$. Together: no real $x$ satisfies both — the system is inconsistent, which is worth checking whenever two quadratic inequalities are imposed at once.