Question

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

Answer 1

$9x-20\ge 0$

$x-\frac{20}{9}\ge 0$

$xϵ[\frac{20}{9},\infty )$

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

Squarring both sides

$9x-20<x^2$

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

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

$xϵ(-\infty ,4)\cup (5,\infty )$

$xϵ[\frac{20}{9},4)\cup (5,\infty )$