Question

Solve the following inequality:
$\sqrt{x^2-4x}>x-3.$

Answer 1

$\sqrt{x^2-4x}>x-3$ - Equation 1

For determination of square roots,

$x^2-4x\ge 0$

$\Rightarrow (x-4)\left(x\right)\ge 0$ - Equation 2

Squaring both sides of the Equation 1,

$x^2-4x>(x-3{)}^2$

$x^2-4x>x^2+9-6x$

$2x>9$

$x>\frac{9}{2}$ - Equation 3

From the figure, range is $x\in [4,\infty )$