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  4. The set of all real numbers satisfying the inequation x2 - |x + 2| + x > 0 is

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Q. The set of all real numbers satisfying the inequation $x^2 - |x + 2| + x > 0 $ is

AP EAMCETAP EAMCET 2019

Solution:

$ x^{2}-|x+2|+x > 0$
Case $I$ If $x+2 \geq 0$
$x^{2}-(x+2)+x > 0 \left\{\because |x| = \begin{cases}
x, & x \geq 0 \\
-x, & x<0
\end{cases}\right\}$
$\Rightarrow x^{2}-2>0 $
$\Rightarrow (x-\sqrt{2})(x+\sqrt{2})=0 $
image
$ x \in[-2,-\sqrt{2}] \cup[\sqrt{2}, \infty]$
Case II $ x + 2 < 0$
$ x^{2}+ (x+2)+x > 0 $
$\Rightarrow x^{2}+2 x+2 > 0$
$\Rightarrow (x+1)^{2}+1 > 0$
$ \Rightarrow x < -2$
Hence, the solution set is
$x \in(-\infty,-\sqrt{2}) \cup(\sqrt{2}, \infty)$