Animated Solution for Mathematics - Quadratic Equations: The set of all real numbers $x$ for which $x^2-|x+2|+x>0$, is

The inequality $x^2-|x+2|+x>0$ contains an absolute value term that changes behavior at the critical point $x=-2$. To solve this, we must partition the number line into two distinct regions based on the sign of the expression inside the modulus.

The Right Side ($x\ge -2$)

In the region where $x\ge -2$, the expression $x+2$ is non-negative. By the definition of the modulus, $|x+2|=x+2$.

Substituting this into the inequality, we obtain:

Simplifying the expression:

The roots of $x^2-2=0$ are $x=\pm \sqrt{2}$. Since the parabola $y=x^2-2$ opens upwards, the inequality holds for $x<-\sqrt{2}$ or $x>\sqrt{2}$.

Applying the boundary condition $x\ge -2$, we find the intersection:

The Left Side ($x<-2$)

Now, we consider the region where $x<-2$. Here, $x+2$ is negative, so the modulus negates the expression: $|x+2|=-(x+2)$.

Substituting this into the original inequality:

To analyze this quadratic, we calculate the discriminant $D$:

Since $D<0$ and the leading coefficient is positive, the quadratic $x^2+2x+2$ is always positive for all real $x$. Given our condition $x<-2$, the solution for this case is:

The Grand Synthesis

We now combine the solutions from both cases using the union operator. From Case 1, we have $[-2,-\sqrt{2})\cup (\sqrt{2},\infty )$, and from Case 2, we have $(-\infty ,-2)$.

The point $x=-2$ acts as a bridge, connecting $(-\infty ,-2)$ and $