Question

$\frac{15-4x}{x^2-x-12} < 4$

Answer 1

$(-\infty, -\frac{3\sqrt{7}}{2}) \cup (-3, \frac{3\sqrt{7}}{2}) \cup (4, \infty)$

Answer 2

To solve the inequality $\frac{15-4x}{x^2-x-12} < 4$:

1. Move all terms to one side: $\frac{15-4x}{x^2-x-12} - 4 < 0$

2. Find a common denominator: $\frac{15-4x - 4(x^2-x-12)}{x^2-x-12} < 0$

3. Expand and simplify the numerator: $\frac{15-4x - 4x^2 + 4x + 48}{x^2-x-12} < 0$ $\frac{-4x^2 + 63}{x^2-x-12} < 0$

4. Multiply by -1 to make the leading coefficient positive (and reverse the inequality): $\frac{4x^2 - 63}{x^2-x-12} > 0$

5. Find the roots of the numerator and denominator:

   • Numerator: $4x^2 - 63 = 0 \implies x = \pm \frac{3\sqrt{7}}{2} \approx \pm 3.969$
   • Denominator: $x^2 - x - 12 = 0 \implies (x-4)(x+3) = 0 \implies x = 4, -3$

6. The critical points are $-\frac{3\sqrt{7}}{2} \approx -3.969$, $-3$, $\frac{3\sqrt{7}}{2} \approx 3.969$, and $4$.

7. Test intervals: $(-\infty, -\frac{3\sqrt{7}}{2})$, $(-\frac{3\sqrt{7}}{2}, -3)$, $(-3, \frac{3\sqrt{7}}{2})$, $(\frac{3\sqrt{7}}{2}, 4)$, $(4, \infty)$.

8. Determine the sign of the expression in each interval:

   • $(-\infty, -\frac{3\sqrt{7}}{2})$: Positive
   • $(-\frac{3\sqrt{7}}{2}, -3)$: Negative
   • $(-3, \frac{3\sqrt{7}}{2})$: Positive
   • $(\frac{3\sqrt{7}}{2}, 4)$: Negative
   • $(4, \infty)$: Positive

9. Since we want the expression to be greater than 0, the solution is $(-\infty, -\frac{3\sqrt{7}}{2}) \cup (-3, \frac{3\sqrt{7}}{2}) \cup (4, \infty)$.