Question:

Step 1: Analyze the first equation. The given equation is $|x + 2| + |x - 2| - 2 = 0$. This can be rewritten as $|x + 2| + |x - 2| = 2$. The expression $|x + 2| + |x - 2|$ represents the sum of the distances from a point $x$ on the number line to the points $-2$ and $2$. The distance between the points $-2$ and $2$ is $|2 - (-2)| = 4$. For any real number $x$, the sum of the distances from $x$ to two distinct points is always greater than or equal to the distance between those two points. Thus, $|x + 2| + |x - 2| \ge 4$. Since $2 < 4$, the equation $|x + 2| + |x - 2| = 2$ has no real solutions. Therefore, the sum of the squares of the roots of the first equation is $0$. Step 2: Analyze the second equation. The given equation is $2x^2 - 2|x - 3| - 5 = 0$. We consider two cases based on the sign of the term $(x - 3)$. Case 1: $x - 3 \ge 0 \Rightarrow x \ge 3$. In this case, $|x - 3| = x - 3$. Substituting this into the equation: $$2x^2 - 2(x - 3) - 5 = 0$$ $$2x^2 - 2x + 6 - 5 = 0$$ $$2x^2 - 2x + 1 = 0$$ To find the roots, we calculate the discriminant $D = b^2 - 4ac$: $$D = (-2)^2 - 4(2)(1) = 4 - 8 = -4$$ Since the discriminant $D < 0$, there are no real roots in this case. Case 2: $x - 3 < 0 \Rightarrow x < 3$. In this case, $|x - 3| = -(x - 3)$. Substituting this into the equation: $$2x^2 - 2(-(x - 3)) - 5 = 0$$ $$2x^2 + 2(x - 3) - 5 = 0$$ $$2x^2 + 2x - 6 - 5 = 0$$ $$2x^2 + 2x - 11 = 0$$ Let the roots of this quadratic equation be $\alpha$ and $\beta$. According to Vieta's formulas, the sum of the roots is $\alpha + \beta = -\frac{2}{2} = -1$. The product of the roots is $\alpha\beta = -\frac{11}{2}$. We must verify that these roots satisfy the condition $x < 3$. The roots are given by the quadratic formula: $$x = \frac{-2 \pm \sqrt{2^2 - 4(2)(-11)}}{2(2)} = \frac{-2 \pm \sqrt{4 + 88}}{4} = \frac{-2 \pm \sqrt{92}}{4} = \frac{-2 \pm 2\sqrt{23}}{4} = \frac{-1 \pm \sqrt{23}}{2}$$ Since $4 < \sqrt{23} < 5$ (as $\sqrt{16} < \sqrt{23} < \sqrt{25}$), we can approximate the roots: $x_1 = \frac{-1 + \sqrt{23}}{2} \approx \frac{-1 + 4.8}{2} = 1.9$. This root satisfies $x < 3$. $x_2 = \frac{-1 - \sqrt{23}}{2} \approx \frac{-1 - 4.8}{2} = -2.9$. This root also satisfies $x < 3$. Both roots are valid. The sum of the squares of the roots is given by the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$. Substituting the values from Vieta's formulas: $$\alpha^2 + \beta^2 = (-1)^2 - 2\left(-\frac{11}{2}\right) = 1 + 11 = 12$$ Thus, the sum of the squares of the roots of the second equation is $12$. Step 3: Calculate the total sum of the squares of the roots. The sum of the squares of the roots of the first equation is $0$. The sum of the squares of the roots of the second equation is $12$. The total sum is $0 + 12 = 12$.