The set of all real numbers satisfying the in | vedclass

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(C) Given inequation: $x^2 - |x+2| + x > 0$
Case $I$: If $x+2 \geq 0$ (i.e., $x \geq -2$), then $|x+2| = x+2$.
The inequation becomes:
$x^2 - (x

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given inequation: $x^2 - |x+2| + x > 0$ Case $I$: If $x+2 \geq 0$ (i.e., $x \geq -2$), then $|x+2| = x+2$. The inequation becomes: $x^2 - (x+2) + x > 0$ $\Rightarrow x^2 - 2 > 0$ $\Rightarrow (x-\sqrt{2})(x+\sqrt{2}) > 0$ The solution for this is $x \in (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)$. Considering the condition $x \geq -2$, the intersection is $x \in [-2, -\sqrt{2}) \cup (\sqrt{2}, \infty)$. Case $II$: If $x+2 < 0$ (i.e., $x < -2$), then $|x+2| = -(x+2)$. The inequation becomes: $x^2 - (-(x+2)) + x > 0$ $\Rightarrow x^2 + x + 2 + x > 0$ $\Rightarrow x^2 + 2x + 2 > 0$ $\Rightarrow (x+1)^2 + 1 > 0$ Since $(x+1)^2 + 1$ is always positive for all real $x$, the condition $x < -2$ is satisfied. Combining Case $I$ and Case $II$: $(-\infty, -2) \cup [-2, -\sqrt{2}) \cup (\sqrt{2}, \infty) = (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)$

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

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