Question

Solve the following inequality:
$\sqrt{x^2-3x-10}<8-x.$

Answer 1

$x^2-3x-10\ge 0$

$(x-5)(x+2)\ge 0$

$xϵ(-\infty ,-2]\cup [5,\infty )$

$\sqrt{x^2-3x-10}<8-x$

Squarring both sides

$x^2-3x-10<x^2-16x+64$

$13x-74<0$

$xϵ(-\infty ,\frac{74}{13})$

$xϵ(-\infty ,-2]\cup [5,\frac{74}{13})$