Question

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

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

Answer 1

The correct option is D

$(-\infty ,-\sqrt{2})\cup (\sqrt{2},\infty )$

Case $1$: When $x+2\ge 0$ i.e $x\ge -2$,

Then given inequality becomes

$x^2-(x+2)+x>0$ ⇒ $x^2-2>0$ ⇒ $|x|>\sqrt{2}$

⇒ Therefore, in this case the part of the solution set is

$[-2,-\sqrt{2})\cup (\sqrt{2},\infty )$.

Case $2$: When $x+2<0$ i.e. $x<-2$,

Then given inequality becomes $x^2+(x+2)+x>0$

⇒ $x^2+2x+2>0$ ⇒ $(x+1{)}^2+1>0$, which is true for all real $x$.

Hence, the part of the solution set in this case is $(-\infty ,-2)$.

Combining the two cases, the solution set is $(-\infty ,-2)\cup [-2,-\sqrt{2})\cup (\sqrt{2},,\infty )=(-\infty ,-\sqrt{2})\cup (\sqrt{2},\infty )$