Key Concepts and Formulas

• Absolute Value: $|x| = x$ if $x \ge 0$, and $|x| = -x$ if $x < 0$.
• Substitution: Simplifying equations by replacing complex expressions with single variables.
• Quadratic Formula (Factoring): Solving quadratic equations of the form $ax^2 + bx + c = 0$.

Step-by-Step Solution

Step 1: Simplify the equation using substitution

• Why: Substitution simplifies the equation and makes it easier to work with.
• Let $y = \sqrt{x}$. Since $x > 0$, we have $y > 0$. Substituting into the given equation: $|\sqrt{x} - 2| + \sqrt{x}(\sqrt{x} - 4) + 2 = 0$ $|y - 2| + y(y - 4) + 2 = 0$ $|y - 2| + y^2 - 4y + 2 = 0$

Step 2: Consider the case $y \ge 2$

• Why: We need to consider the cases where the expression inside the absolute value is non-negative.
• If $y \ge 2$, then $|y - 2| = y - 2$. Substituting into the equation: $(y - 2) + y^2 - 4y + 2 = 0$ $y^2 - 3y = 0$ $y(y - 3) = 0$ This yields $y = 0$ or $y = 3$.

Step 3: Validate the solutions for $y \ge 2$

• Why: We need to check if the solutions obtained satisfy the condition $y \ge 2$.
• Since we assumed $y \ge 2$, we check if $y = 0$ or $y = 3$ satisfies this condition.
  • $y = 0$ does not satisfy $y \ge 2$, so we reject $y = 0$.
  • $y = 3$ satisfies $y \ge 2$, so $y = 3$ is a valid solution.

Step 4: Convert back to $x$ for the case $y \ge 2$

• Why: We need to find the corresponding value of $x$ for the valid solution of $y$.
• Since $y = \sqrt{x}$, we have $\sqrt{x} = 3$. Squaring both sides gives $x = 9$.

Step 5: Consider the case $y < 2$

• Why: We need to consider the cases where the expression inside the absolute value is negative.
• If $y < 2$, then $|y - 2| = -(y - 2) = 2 - y$. Substituting into the equation: $(2 - y) + y^2 - 4y + 2 = 0$ $y^2 - 5y + 4 = 0$ $(y - 1)(y - 4) = 0$ This yields $y = 1$ or $y = 4$.

Step 6: Validate the solutions for $y < 2$

• Why: We need to check if the solutions obtained satisfy the condition $y < 2$.
• Since we assumed $y < 2$, we check if $y = 1$ or $y = 4$ satisfies this condition.
  • $y = 1$ satisfies $y < 2$, so $y = 1$ is a valid solution.
  • $y = 4$ does not satisfy $y < 2$, so we reject $y = 4$.

Step 7: Convert back to $x$ for the case $y < 2$

• Why: We need to find the corresponding value of $x$ for the valid solution of $y$.
• Since $y = \sqrt{x}$, we have $\sqrt{x} = 1$. Squaring both sides gives $x = 1$.

Step 8: Calculate the sum of the solutions

• Why: The problem asks for the sum of all valid solutions for $x$.
• The valid solutions are $x = 9$ and $x = 1$. The sum of the solutions is $9 + 1 = 10$.

Common Mistakes & Tips

• Forgetting to validate solutions: Always check if the solutions obtained satisfy the conditions of the case being considered.
• Incorrectly applying the absolute value: Make sure to correctly determine the sign of the expression inside the absolute value before removing the absolute value signs.
• Not checking the domain: Remember that $\sqrt{x}$ is only defined for $x \ge 0$, and in this case, $x > 0$.

Summary

By substituting $y = \sqrt{x}$, we simplified the equation and solved it by considering two cases based on the sign of $y - 2$. We validated the solutions obtained in each case and converted them back to $x$. Finally, we calculated the sum of the valid solutions for $x$, which is 10.

Final Answer

The final answer is \boxed{10}, which corresponds to option (D).