Question

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

Answer 1

$\frac{15-4x}{x^2-x-12}<4\frac{15-4x}{x^2-x-12}-4<0\frac{15-4x-{4x}^2+4x+48}{(x-4)(x+3)}<0\frac{{4x}^2-63}{(x-4)(x+3)}>0\frac{(x-\sqrt{\frac{63}{4}})(x+\sqrt{\frac{63}{4}})}{(x-4)(x+3)}>0x\in (-\infty ,-\sqrt{\frac{63}{4}})\cup (-3,\sqrt{\frac{63}{4}})\cup (4,\infty )$