Sum of Squares of Roots for Absolute Value Equations

The problem requires finding the sum of the squares of the roots for two distinct equations involving absolute values.

Solving Equation 1: $|x-2|^2+|x-2|-2=0$

1. Let $y = |x-2|$. The equation transforms into a quadratic equation in terms of $y$:

   $y^2 + y - 2 = 0$

2. Factor the quadratic equation:

   $(y+2)(y-1) = 0$

   This gives potential values for $y$: $y=-2$ or $y=1$.

3. Since $y = |x-2|$, $y$ must be non-negative ($y \ge 0$). Therefore, we discard $y=-2$.

   We have $|x-2| = 1$.

4. Solve for $x$:

   This implies either $x-2 = 1$ or $x-2 = -1$.

   • $x-2 = 1 \implies x = 3$
   • $x-2 = -1 \implies x = 1$
5. The roots of the first equation are $1$ and $3$.
6. Calculate the sum of the squares of these roots:

   Sum of squares = $1^2 + 3^2 = 1 + 9 = 10$.

Solving Equation 2: $x^2-2|x-3|-5=0$

We need to consider two cases based on the sign of $(x-3)$.

Case 1: $x \ge 3$

1. In this case, $|x-3| = x-3$. Substitute this into the equation:

   $x^2 - 2(x-3) - 5 = 0$

2. Simplify and solve the resulting quadratic equation:

   $x^2 - 2x + 6 - 5 = 0$

   $x^2 - 2x + 1 = 0$

   $(x-1)^2 = 0$

   This yields $x=1$.

3. Check the condition: The solution $x=1$ does not satisfy the condition $x \ge 3$. Therefore, there are no roots from this case.

Case 2: $x < 3$

1. In this case, $|x-3| = -(x-3) = 3-x$. Substitute this into the equation:

   $x^2 - 2(3-x) - 5 = 0$

2. Simplify and solve the resulting quadratic equation:

   $x^2 - 6 + 2x - 5 = 0$

   $x^2 + 2x - 11 = 0$

3. Find the roots using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:

   $x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-11)}}{2(1)}$

   $x = \frac{-2 \pm \sqrt{4 + 44}}{2}$

   $x = \frac{-2 \pm \sqrt{48}}{2}$

   $x = \frac{-2 \pm 4\sqrt{3}}{2}$

   $x = -1 \pm 2\sqrt{3}$

4. Check the condition:
   • $x = -1 + 2\sqrt{3} \approx -1 + 2(1.732) = 2.464$. This satisfies $x < 3$.
   • $x = -1 - 2\sqrt{3} \approx -1 - 3.464 = -4.464$. This also satisfies $x < 3$.
5. The roots of the second equation are $-1 + 2\sqrt{3}$ and $-1 - 2\sqrt{3}$.
6. Calculate the sum of the squares of these roots. For the equation $x^2 + 2x - 11 = 0$, let the roots be $x_3$ and $x_4$. Using Vieta's formulas:
   • Sum of roots: $x_3 + x_4 = -\frac{2}{1} = -2$.
   • Product of roots: $x_3 x_4 = \frac{-11}{1} = -11$.
   The sum of squares is given by $(x_3 + x_4)^2 - 2x_3 x_4$:

   Sum of squares = $(-2)^2 - 2(-11) = 4 + 22 = 26$.

Calculating the Total Sum of Squares

Add the sum of the squares of the roots from both equations:

Total Sum = (Sum of squares from Eq 1) + (Sum of squares from Eq 2)

Total Sum = $10 + 26 = 36$.