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: $|x-2|^2+|x-2|-2=0$
- Let $y = |x-2|$. The equation transforms into a quadratic equation in terms of $y$:
$y^2 + y - 2 = 0$
- Factor the quadratic equation:
$(y+2)(y-1) = 0$
This gives potential values for $y$: $y=-2$ or $y=1$.
- Since $y = |x-2|$, $y$ must be non-negative ($y \ge 0$). Therefore, we discard $y=-2$.
We have $|x-2| = 1$.
- Solve for $x$:
This implies either $x-2 = 1$ or $x-2 = -1$.
- $x-2 = 1 \implies x = 3$
- $x-2 = -1 \implies x = 1$
- The roots of the first equation are $1$ and $3$.
- 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: $x \ge 3$
- In this case, $|x-3| = x-3$. Substitute this into the equation:
$x^2 - 2(x-3) - 5 = 0$
- Simplify and solve the resulting quadratic equation:
$x^2 - 2x + 6 - 5 = 0$
$x^2 - 2x + 1 = 0$
$(x-1)^2 = 0$
This yields $x=1$.
- 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: $x < 3$
- In this case, $|x-3| = -(x-3) = 3-x$. Substitute this into the equation:
$x^2 - 2(3-x) - 5 = 0$
- Simplify and solve the resulting quadratic equation:
$x^2 - 6 + 2x - 5 = 0$
$x^2 + 2x - 11 = 0$
- Find the roots using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$:
$x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-11)}}{2(1)}$
$x = \frac{-2 \pm \sqrt{4 + 44}}{2}$
$x = \frac{-2 \pm \sqrt{48}}{2}$
$x = \frac{-2 \pm 4\sqrt{3}}{2}$
$x = -1 \pm 2\sqrt{3}$
- 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 $x < 3$.
- The roots of the second equation are $-1 + 2\sqrt{3}$ and $-1 - 2\sqrt{3}$.
- 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: $x_3 x_4 = \frac{-11}{1} = -11$.
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 = $10 + 26 = 36$.
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