Question

Linear and Quadralic Equations and Inequalities.
Solve the following inequalities.
$2x^2-5|x|+3⩾0.$

Answer 1

$Case-I:x\ge 0$

$2x^2-5|x|+3\ge 0$

$⟹2x^2-5x+3\ge 0$

$⟹(2x-3)(x-1)\ge 0$

$⟹x\le 1$ or $x\ge \frac{3}{2}$ that is $x\in [0,1]\cup [\frac{3}{2},\infty ]$

This is the solution for $+ve$ x.

$Case-I:x\le 0$

$2x^2-5|x|+3\ge 0$

$⟹2x^2+5x+3\ge 0$

$⟹(2x+3)(x+1)\ge 0$

$⟹x\le \frac{-3}{2}$ or $x\ge -1$ that is $x\in [-\infty ,\frac{-3}{2}]\cup [-1,0]$

This is the solution for $-ve$ x.

Hence solution is the union of these two solutions.

$x\in [-\infty ,\frac{-3}{2}]\cup [-1,1]\cup [\frac{3}{2},\infty ]$