The set of all real numbers x for which {x^2} | vedclass

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(B) Case $I$: When $x + 2 \ge 0$ i.e.,$x \ge -2,$The given inequality becomes ${x^2} - (x + 2) + x > 0 \implies {x^2} - 2 > 0 \implies |x| > \sqrt{2}.

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

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(B) Case $I$: When $x + 2 \ge 0$ i.e.,$x \ge -2,$The given inequality becomes ${x^2} - (x + 2) + x > 0 \implies {x^2} - 2 > 0 \implies |x| > \sqrt{2}.$This implies $x  \sqrt{2}.$Since $x \ge -2,$ the solution set for this case is $[ -2, -\sqrt{2} ) \cup (\sqrt{2}, \infty ).$Case $II$: When $x + 2 The given inequality becomes ${x^2} + (x + 2) + x > 0 \implies {x^2} + 2x + 2 > 0.$Since the discriminant $D = 2^2 - 4(1)(2) = 4 - 8 = -4 Thus,the solution set for this case is $( -\infty, -2).$Combining the two cases,the total solution set is $( -\infty, -2) \cup [ -2, -\sqrt{2} ) \cup (\sqrt{2}, \infty ) = ( -\infty, -\sqrt{2} ) \cup (\sqrt{2}, \infty ).$

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

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