Question

The set of all real numbers x for which $x^2-|x+2|+x>0$ is $1$

Answer 1

The condition given in the question is $x^2–|x+2|+x>0$

Two cases are possible:

Case 1: When$(x+2)\ge 0.$

Therefore, $x^2–x–2+x>0$

$x^2–2>0$

So, either $x=+\sqrt{2}$or $-\sqrt{2}$

Hence, $x\in [-2,-\surd 2)\cup (\surd 2,\infty )$……. (1)

Case 2: When $(x+2)<0$

Then, $x^2+x+2+x>0$

So, $x^2+2x+2>0$

This gives ${(x+1)}^2+1>0$ and this is true for every $x$

Hence,$x\le -2orx\in (-\infty ,-2)$ ……… (2)

Hence, From equations (2) and (3) we get $x\in (-\infty ,-\surd 2)\cup (\surd 2,\infty )$ set of all real numbers.