Question

Find the set of all $x$ for which$\frac{2x}{2x^2+5x+2}>\frac{1}{x+1}$

• A. $xϵ(-2,-1)\cup (\frac{-2}{3},\frac{-1}{2})$
• B. $-2\ge x\ge -1$
• C. $-2\ge x<-1$
• D. $-2<x\le -1$

Answer 1

Answer: (A)

$xϵ(-2,-1)\cup (\frac{-2}{3},\frac{-1}{2})$

$\frac{2x}{x^2+5x+2}>\frac{1}{x+1}$

$\frac{2x}{{2x}^2+4x+x+2}>\frac{1}{x+1}$

$\frac{2x}{(2x+1)(x+2)}-\frac{1}{x+1}>0$

$\frac{{2x}^2+2x-{2x}^2-5x-2}{(2x+1)(x+2)(x+1)}>0$

$\frac{-3x-2}{(2x+1)(x+2)(x+1)}>0$

$\frac{3x+2}{(2x+1)(x+2)(x+1)}<0$

$x\epsilon (-2,-1)\cup (\frac{-2}{3},\frac{-1}{2})$