Question

Values of $x$ simultaneously satisfying: $x^2 - x - 6 \ge 0$ & $x^2 - 4x < 0$

Answer 1

$x \in [3, 4)$

Answer 2

The problem requires finding the intersection of solution sets for two inequalities.

1. For $x^2 - x - 6 \ge 0$: The roots of $x^2 - x - 6 = 0$ are found using the quadratic formula or factoring. Factoring gives $(x-3)(x+2)=0$, so the roots are $x=3$ and $x=-2$. Since this is a parabola opening upwards, the inequality $x^2 - x - 6 \ge 0$ holds for values of $x$ outside the roots. Thus, the solution set is $x \in (-\infty, -2] \cup [3, \infty)$.

2. For $x^2 - 4x < 0$: The roots of $x^2 - 4x = 0$ are found by factoring: $x(x-4)=0$, so the roots are $x=0$ and $x=4$. Since this is a parabola opening upwards, the inequality $x^2 - 4x < 0$ holds for values of $x$ between the roots. Thus, the solution set is $x \in (0, 4)$.

3. Intersection of the solution sets: We need to find the values of $x$ that are in both $(-\infty, -2] \cup [3, \infty)$ and $(0, 4)$.

   • The interval $(-\infty, -2]$ does not overlap with $(0, 4)$.
   • The interval $[3, \infty)$ overlaps with $(0, 4)$ in the interval $[3, 4)$.

   Therefore, the intersection of the two solution sets is $[3, 4)$.