Question

Solve the inequation
$\sqrt{x^2-3x-10}>(8-x)$

Answer 1

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

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

$x\in (-\infty ,-2)\cup (5,\infty )$

$\sqrt{x^2-3x-10}>(8-x)$

squaring both sides

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

$13x-74>0$

$x>\frac{74}{13}$

$x\in (\frac{74}{13},\infty )$

$\therefore x\in (\frac{74}{13},\infty )$