Question

Solve: $\frac{x^2+5x+3}{x+2}<x$

Answer 1

$\frac{x^2+5x+3}{x+2}<x$
$\Rightarrow \frac{x^2+5x+3}{x+2}-x<0$
$\Rightarrow \frac{x^2+5x+3-x^2-2x}{x+2}<0\Rightarrow \frac{3x+3}{x+2}<0$
$\Rightarrow \frac{x+1}{x+2}<0$ ......(1)
To solve (1) it is sufficient to solve $(x+1)(x+2)<0$
We know that $(x-\alpha )(x-\beta )<0$
$\Rightarrow \alpha <x<\beta$ where $(\alpha <\beta )$
$\therefore -2<x<-1$
Thus the required solution is $-2<x<-1$