Question

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

Let $f\left(x\right)=x^2-|x+2|+x>0$

If $x\le -2$

$f\left(x\right)=x^2+x+2+x$

$=x^2+2x+2$

$△=2^2-4\left(2\right)$

$△=-4<0$

$\therefore f\left(x\right)=x^2+2x+2>0$ (always)

$xϵ[-\infty ,-2]\to 1$

If $x>-2$

$f\left(x\right)=x^2-x-2+x>0$

$x^2-2>0$

$\therefore xϵ(-\infty ,-\sqrt{2})\bigcup (\sqrt{2},\infty )$

$xϵ(-\infty ,-\sqrt{2})\bigcup (\sqrt{2},\infty )\to 2$

From $1$ and $2,xϵ(-\infty ,-\sqrt{2})\bigcup (\sqrt{2},\infty )$

$\therefore xϵ(-\infty ,-\sqrt{2})\bigcup (\sqrt{2},\infty )$