Question

For what real x, can we have the inequality I4x-x²-3I / √(x+1) >0

Answer 1

Solution:

$\dfrac{\left|4x-x^2-3\right|}{\sqrt{x+1}}>0 \\ \Rightarrow\dfrac{\left|-x^2+4x-3\right|}{\sqrt{x+1}}>0 \\ \Rightarrow\dfrac{\left|-x^2+3x+x-3\right|}{\sqrt{x+1}}>0 \\ \Rightarrow\dfrac{\left|(-x+1)(x-3)\right|}{\sqrt{x+1}}>0$

Identifying the intervals: $-1<x<1\quad \mathrm{or}\quad \:1<x<3\quad \mathrm{or}\quad \:x>3$

Finding non-negative value for radical: $x\ge-1$

Merging overlapping intervals, the solution is $\left(-1,\:1\right)\cup \left(1,\:3\right)\cup \left(3,\:\infty \:\right)$