Question

The set of all real number $x$ 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 $(-\infty ,-\sqrt{2})\cup (\sqrt{2},\infty )$

$x^2-|x+2|+x>0$
$x^2-(x+2)+x>0$ and $x^2+(x+2)+x>0$
$\Rightarrow x^2-2>0$ and $x^2+2x+2>0$
$\Rightarrow (x-\sqrt{2})(x+\sqrt{2})>0$ and $(x+1{)}^2>-1$
$\Rightarrow x\in (-\infty ,-\sqrt{2})\cup (\sqrt{2},\infty )$ and no real value.
Hence, $x\in (-\infty ,-\sqrt{2})\cup (\sqrt{2},\infty )$.