Question

The set of all real $x$ for which $x^2-|x+2|+x>0$ is

• A. $(-\infty ,-2)\cup (2,\infty )$
• B. $(-\infty ,-\sqrt{2})\cup (\sqrt{2},\infty )$
• C. $(-\infty ,-1)\cup (1,\infty )$
• D. $(\sqrt{2},\infty )$

Answer 1

The correct option is B $(-\infty ,-\sqrt{2})\cup (\sqrt{2},\infty )$

Case 1:
$x+2\ge 0\Rightarrow x\ge -2$
The equation now becomes $x^2-2\ge 0\Rightarrow x\in [-\infty ,-\sqrt{2}],[\sqrt{2},\infty ],\Rightarrow x\in [-2,-\sqrt{2}],[\sqrt{2},\infty ]$
Case 2:
$x+2<0\Rightarrow x<-2$
The equation now becomes $x^2+2x+2<0\Rightarrow (x+1{)}^2+1<0\Rightarrow x\in \varphi$
Hence $x\in [-2,-\sqrt{2}],[\sqrt{2},\infty ]$