Key Concepts and Formulas

• Absolute Value Definition: $|x| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases}$
• Quadratic Formula: For a quadratic equation $ax^2 + bx + c = 0$, the roots are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Squaring a binomial: $(a \pm b)^2 = a^2 \pm 2ab + b^2$

Step-by-Step Solution

Step 1: Define Cases based on the Absolute Value

The given equation is $x^2 + |2x - 3| - 4 = 0$. We need to consider two cases based on the sign of the expression inside the absolute value, which is $2x - 3$. We find the critical point by setting $2x-3 = 0$, which gives $x = \frac{3}{2}$.

Step 2: Case 1: $2x - 3 \geq 0 \implies x \geq \frac{3}{2}$

If $x \geq \frac{3}{2}$, then $|2x - 3| = 2x - 3$. Substituting this into the equation, we get: $x^2 + (2x - 3) - 4 = 0$ $x^2 + 2x - 7 = 0$ Now, we use the quadratic formula to solve for $x$: $x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-7)}}{2(1)}$ $x = \frac{-2 \pm \sqrt{4 + 28}}{2}$ $x = \frac{-2 \pm \sqrt{32}}{2}$ $x = \frac{-2 \pm 4\sqrt{2}}{2}$ $x = -1 \pm 2\sqrt{2}$ We have two potential roots: $x_A = -1 + 2\sqrt{2}$ and $x_B = -1 - 2\sqrt{2}$.

Now we check if these roots satisfy the condition $x \geq \frac{3}{2} = 1.5$.

• $x_A = -1 + 2\sqrt{2} \approx -1 + 2(1.414) = -1 + 2.828 = 1.828$. Since $1.828 \geq 1.5$, $x_A$ is a valid root.
• $x_B = -1 - 2\sqrt{2} \approx -1 - 2(1.414) = -1 - 2.828 = -3.828$. Since $-3.828 < 1.5$, $x_B$ is not a valid root.

Thus, in Case 1, we have one valid root: $x_1 = -1 + 2\sqrt{2}$.

Step 3: Case 2: $2x - 3 < 0 \implies x < \frac{3}{2}$

If $x < \frac{3}{2}$, then $|2x - 3| = -(2x - 3) = -2x + 3$. Substituting this into the equation, we get: $x^2 + (-2x + 3) - 4 = 0$ $x^2 - 2x - 1 = 0$ Using the quadratic formula to solve for $x$: $x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)}$ $x = \frac{2 \pm \sqrt{4 + 4}}{2}$ $x = \frac{2 \pm \sqrt{8}}{2}$ $x = \frac{2 \pm 2\sqrt{2}}{2}$ $x = 1 \pm \sqrt{2}$ We have two potential roots: $x_C = 1 + \sqrt{2}$ and $x_D = 1 - \sqrt{2}$.

Now we check if these roots satisfy the condition $x < \frac{3}{2} = 1.5$.

• $x_C = 1 + \sqrt{2} \approx 1 + 1.414 = 2.414$. Since $2.414 \nless 1.5$, $x_C$ is not a valid root.
• $x_D = 1 - \sqrt{2} \approx 1 - 1.414 = -0.414$. Since $-0.414 < 1.5$, $x_D$ is a valid root.

Thus, in Case 2, we have one valid root: $x_2 = 1 - \sqrt{2}$.

Step 4: Calculate the Sum of the Squares of the Valid Roots

We need to find $x_1^2 + x_2^2$, where $x_1 = -1 + 2\sqrt{2}$ and $x_2 = 1 - \sqrt{2}$.

$x_1^2 = (-1 + 2\sqrt{2})^2 = (-1)^2 + 2(-1)(2\sqrt{2}) + (2\sqrt{2})^2 = 1 - 4\sqrt{2} + 8 = 9 - 4\sqrt{2}$ $x_2^2 = (1 - \sqrt{2})^2 = (1)^2 + 2(1)(-\sqrt{2}) + (-\sqrt{2})^2 = 1 - 2\sqrt{2} + 2 = 3 - 2\sqrt{2}$

Therefore, $x_1^2 + x_2^2 = (9 - 4\sqrt{2}) + (3 - 2\sqrt{2}) = 12 - 6\sqrt{2} = 6(2 - \sqrt{2})$

Common Mistakes & Tips

• Forgetting to Check the Roots: A common mistake is to forget to check if the solutions obtained in each case satisfy the initial condition for that case (e.g., $x \geq \frac{3}{2}$ or $x < \frac{3}{2}$).
• Sign Errors: Be careful with signs, especially when dealing with the absolute value and the quadratic formula.
• Simplifying Radicals: Always simplify radicals to their simplest form.

Summary

We split the absolute value equation into two cases based on the sign of the expression inside the absolute value. We solved the resulting quadratic equations and checked if the solutions satisfied the conditions for each case. Finally, we calculated the sum of the squares of the valid roots, which is $6(2 - \sqrt{2})$.

Final Answer

The final answer is \boxed{6(2-\sqrt{2})}, which corresponds to option (A).