Question

Solve the following inequality:
$\sqrt{9x-20}>x.$

Answer 1

$9x-20\ge 0$

$xϵ[\frac{20}{9},\infty )$

$\sqrt{9x-20}>x$

Squarring both sides

$9x-20>x^2$

$x^2-9x+20<0$

$(x-5)(x-4)<0$

$xϵ(4,5)$