Question

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

Answer 1

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

Answer 2

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

1. Move all terms to one side:

   $\frac{15-4x}{x^{2}-x-12} - 4 < 0$

2. Combine into a single fraction:

   $\frac{15-4x - 4(x^{2}-x-12)}{x^{2}-x-12} < 0$

   $\frac{15-4x - 4x^{2}+4x+48}{x^{2}-x-12} < 0$

   $\frac{-4x^{2} + 63}{x^{2}-x-12} < 0$

3. Multiply by -1 and reverse the inequality:

   $\frac{4x^{2} - 63}{x^{2}-x-12} > 0$

4. Factor the numerator and denominator:

   Numerator: $4x^{2} - 63 = (2x - 3\sqrt{7})(2x + 3\sqrt{7})$

   Denominator: $x^{2}-x-12 = (x-4)(x+3)$

   So the inequality becomes:

   $\frac{(2x - 3\sqrt{7})(2x + 3\sqrt{7})}{(x-4)(x+3)} > 0$

5. Find critical points:

   Numerator: $x = \pm \frac{3\sqrt{7}}{2}$

   Denominator: $x = 4, x = -3$

6. Arrange critical points in increasing order:

   $-\frac{3\sqrt{7}}{2} \approx -3.9675, -3, \frac{3\sqrt{7}}{2} \approx 3.9675, 4$

7. 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

8. Identify intervals where the expression is greater than 0:

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

Thus, the solution is $x \in (-\infty, -\frac{3\sqrt{7}}{2}) \cup (-3, \frac{3\sqrt{7}}{2}) \cup (4, \infty)$.