Question

1. $|4x|+x^{2}>0$

Explanation:

Step1: Analyze cases for \(x\)

Case 1: When \(x\geq0\), \(|4x| = 4x\), so the inequality becomes \(4x + x^{2}>0\). Factor it as \(x(4 + x)>0\). The solutions of the quadratic - inequality \(y=x(4 + x)\) (where \(y = 0\) when \(x=0\) or \(x=-4\)) for \(y>0\) and \(x\geq0\) are \(x>0\).

Step2: Analyze the other case

When \(x<0\), \(|4x|=-4x\), and the inequality is \(-4x + x^{2}>0\). Factor it as \(x(x - 4)>0\). The solutions of the quadratic - inequality \(y=x(x - 4)\) (where \(y = 0\) when \(x = 0\) or \(x = 4\)) for \(y>0\) and \(x<0\) are \(x<0\).

Answer:

The solution set is \((-\infty,0)\cup(0,\infty)\)