Question

Solve the following inequality:
$\sqrt{x^2-3x+2}>2x-5.$

Answer 1

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

$(x-2)(x-1)\ge 0$

$xϵ(-\infty ,1]\cup [2,\infty )$

$\sqrt{x^2-3x+2}>2x-5$

Squarring both sides

$x^2-3x+2>4x^2-20x+25$

$3x^2-17x+23<0$

As discriminant less than $0$ So, $3x^2-17x+23>0\forall xϵR$

$xϵ(-\infty ,-1]\cup [2,\infty )$