Question

If $x^2-|x+3|+x>0,$ then $x\in$

• A. $(-\infty ,-\sqrt{3})$
• B. $(-3,\sqrt{3})$
• C. $(-\sqrt{3},\sqrt{3})$
• D. $(\sqrt{3},\infty )$

Answer 1

Answer: (A), (D)

The correct options are
A $(-\infty ,-\sqrt{3})$
D $(\sqrt{3},\infty )$

$x^2-|x+3|+x>0$
If $x>-3$,
then $x^2–3>0$
$\Rightarrow x>\sqrt{3}\text{or}x<-\sqrt{3}$
$\Rightarrow x\in (-\infty ,-\sqrt{3})\cup (\sqrt{3},\infty )$

If $x<-3$,
then $x^2+x+3+x>0$
$\Rightarrow x^2+2x+3>0$
This is true for all $x\in R$ as $D<0$

$\therefore x\in (-\infty ,-\sqrt{3})\cup (\sqrt{3},\infty )$