Key Concepts and Formulas

• Absolute Value: $|x| = x$ if $x \ge 0$, and $|x| = -x$ if $x < 0$.
• Substitution: Replacing a complex expression with a single variable to simplify the equation.
• Solving Quadratic Equations: Factoring, completing the square, or using the quadratic formula to find the roots of a quadratic equation.

Step-by-Step Solution

Step 1: Simplify the equation and introduce a substitution.

The given equation is: $2\left| {\sqrt x - 3} \right| + \sqrt x \left( {\sqrt x - 6} \right) + 6 = 0$ We are given that $x \in \mathbb{R}$ and $x \ge 0$. First, expand the term $\sqrt{x} \left( {\sqrt x - 6} \right)$: $\sqrt x \left( {\sqrt x - 6} \right) = (\sqrt{x})^2 - 6\sqrt{x} = x - 6\sqrt{x}$ Substituting this back into the original equation, we have: $2\left| {\sqrt x - 3} \right| + x - 6\sqrt{x} + 6 = 0$ Let $y = \sqrt{x}$. Since $x \ge 0$, we have $y \ge 0$. Substituting $y$ into the equation gives: $2|y - 3| + y^2 - 6y + 6 = 0$ This substitution simplifies the equation, turning it into a quadratic with an absolute value.

Step 2: Solve the equation by considering cases for the absolute value.

We consider two cases based on the sign of $y - 3$.

• Case 1: $y - 3 \ge 0$ (which implies $y \ge 3$). In this case, $|y - 3| = y - 3$. Substituting this into the equation, we get: $2(y - 3) + y^2 - 6y + 6 = 0$ Expanding and simplifying: $2y - 6 + y^2 - 6y + 6 = 0$ $y^2 - 4y = 0$ Factoring out $y$: $y(y - 4) = 0$ This yields two potential solutions for $y$: $y = 0$ or $y = 4$. We must check these solutions against the condition $y \ge 3$:

  • $y = 0$: This does not satisfy $y \ge 3$, so it is an extraneous solution.
  • $y = 4$: This satisfies $y \ge 3$, so it is a valid solution.

• Case 2: $y - 3 < 0$ (which implies $y < 3$). In this case, $|y - 3| = -(y - 3) = 3 - y$. Substituting this into the equation, we get: $2(3 - y) + y^2 - 6y + 6 = 0$ Expanding and simplifying: $6 - 2y + y^2 - 6y + 6 = 0$ $y^2 - 8y + 12 = 0$ Factoring the quadratic: $(y - 2)(y - 6) = 0$ This yields two potential solutions for $y$: $y = 2$ or $y = 6$. We check these against the condition $y < 3$:

  • $y = 2$: This satisfies $y < 3$, so it is a valid solution.
  • $y = 6$: This does not satisfy $y < 3$, so it is an extraneous solution.

Step 3: Consolidate valid solutions and convert back to $x$.

The valid solutions for $y$ are $y = 4$ and $y = 2$. Both of these also satisfy the initial condition $y \ge 0$. Now we convert back to $x$ using $x = y^2$:

• If $y = 2$: $x = 2^2 = 4$
• If $y = 4$: $x = 4^2 = 16$

Step 4: Verify solutions in the original equation.

We have $x = 4$ and $x = 16$. We check these solutions in the original equation:

• For $x = 4$: $2|\sqrt{4} - 3| + \sqrt{4}(\sqrt{4} - 6) + 6 = 2|2 - 3| + 2(2 - 6) + 6 = 2(1) + 2(-4) + 6 = 2 - 8 + 6 = 0$ This is true.

• For $x = 16$: $2|\sqrt{16} - 3| + \sqrt{16}(\sqrt{16} - 6) + 6 = 2|4 - 3| + 4(4 - 6) + 6 = 2(1) + 4(-2) + 6 = 2 - 8 + 6 = 0$ This is true.

Therefore, the set $S = \{4, 16\}$. The set $S$ contains exactly two elements.

Common Mistakes & Tips

• Missing Cases: Always consider all possible cases when dealing with absolute values.
• Extraneous Solutions: Remember to check solutions against the conditions imposed by the absolute value cases and the original equation.
• Algebraic Errors: Pay close attention to signs and exponents during algebraic manipulations.

Summary

By substituting $y = \sqrt{x}$, considering cases for the absolute value, solving the resulting quadratic equations, and checking for extraneous solutions, we found the valid solutions for $x$. The set $S$ contains two elements, $4$ and $16$.

The final answer is \boxed{contains exactly two elements}, which corresponds to option (D).