Question

Solve the following inequality:
$\sqrt{8+2x-x^2}>6-3x.$

Answer 1

$\sqrt{8+2x-x^2}>6-3x$

Squaring

$8+2x-x^2>36-36x+9x^2$

$10x^2-38x+28<0$

$5x^2-19x+14<0$

$5x^2-14x-5x+14<0$

$(x-1)(5x-14)<0$

$x\in (1,\frac{14}{5})$

$8+2x-x^2\ge 0$

$x^2-2x-8\le 0$

$(x+2)(x-4)\le 0$

$x\in (-2,4)$

$\therefore x\in (1,\frac{14}{5})$