Question

Solve $\frac{6x^2-5x-3}{x^2-2x+6}\le 4$

Answer 1

$\frac{6x^2-5x-3}{x^2-2x+6}\le 4$

$6x^2-5x-3\le 4(x^2-2x+6)$

$6x^2-5x-3\le (4x^2-8x+24)$

$2x^2+3x-27\le 0$

$(x-3)(x-\frac{9}{2})\le 0$

$⟹(x-3)\le 0$ and $(x-\frac{9}{2})\ge 0$ or

$(x-3)\ge 0$ and $(x-\frac{9}{2}\le 0)$

Out of this $(x-3)\ge 0$ and $(x-\frac{9}{2}\le 0)$ is valid.

$⟹x\ge 3$ and $x\le 4.5$