Question

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

Answer 1

$3x-x^2\ge 0$

$x^2-3x\le 0$

$x(x-3)\le 0$

$xϵ[0,3]$

$\sqrt{3x-x^2}<4-x$

Squarring both sides

$3x-x^2<16+x^2-8x$

As discriminant less than zero so $2x^2-11x+16>0$ for every $xϵR$.

$xϵ[0,3]$