Question

Linear and Quadralic Equations and Inequalities.
Solve the following inequalities.
$x^2-5|x|+4<0.$

Answer 1

$Case-I:x\ge 0$

$x^2-5|x|+4<0$

$⟹x^2-5x+4<0$

$⟹(x-1)(x-4)<0$

$⟹1<x<4$ that is $x\in (1,4)$

This is the solution for $+ve$ x.

$Case-I:x\le 0$

$x^2-5|x|+4<0$

$⟹x^2+5x+4<0$

$⟹(x+1)(x+4)<0$

$⟹-4<x<-1$ that is $x\in (-4,-1)$

This is the solution for $-ve$ x.

Hence, solution set is $x\in (-4,-1)\cap (1,4)$