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

Question

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

Smart Achievers · Accepted Answer

(2)

$x^{2} -$ $| x + 2 |$ $+ x > 0$

When $x + 2 > 0$ ,

$Then$ $x^{2} - (x + 2) + x > 0,$ $x^{2} - x - 2 + x > 0 \Rightarrow x^{2} - 2 > 0$

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

$x \in (- 2, - \sqrt{2}) \cup (+ \sqrt{2}, \infty)$ ...(i)

When $x + 2 < 0$ , $| x + 2 |$ $= - (x + 2)$ ,

$\Rightarrow x^{2} + x + 2 + x > 0$ $\forall x \in R$ ...(ii)

From (i) and (ii), we get $x \in (- \infty, - \sqrt{2}) \cup (\sqrt{2}, \infty)$

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