Given \( \Large x^{2}-|x+2|+x>0 \) we know that\( \Large |x+2|=\pm \left(x+2\right) \) If \( \Large |x+2|=-x+2x\ge -2 \) Equation (i) becomes \( \Large \left(when\ x \ge - 2 \right) \) \( \Large x^{2}-x-2+x>0 \) \( \Large x^{2}-2 > 0 \) => \( \Large x < -\sqrt{2}\ or\ x > \sqrt{2} \) \( \Large x \epsilon \left[ -2, -\sqrt{2} \right] \cup \left(\sqrt{2}, \infty\right) \) If therefore, equation (i) becomes \( \Large x^{2}+x+2+x>0\ when\ x \le -2 \) \( \Large x^{2}+2x+2 > 0\ when\ x \le -2 \) \( \Large \left(x+1\right)^{2}+1 > 0\ when x \le -2 \) which is true for all x. \( \Large x < -2\ or\ x \left(-\infty,\ -2\right) \) from equations (ii) and (iii) \( \Large x \epsilon \left(-\infty,\ -\sqrt{2} \right) \cup \left(\sqrt{2},\ \infty\right) \)
