Key Concepts and Formulas

• Substitution: Simplifying equations by replacing complex expressions with a single variable.
• Quadratic Equation Factoring: Solving quadratic equations of the form $ax^2 + bx + c = 0$ by factoring into $(px+q)(rx+s)=0$.
• Domain of Square Root Functions: The expression $\sqrt{x}$ is only defined for non-negative real numbers, i.e., $x \ge 0$. In the denominator, we require $x>0$.

Step-by-Step Solution

Step 1: Identify the Domain and Perform a Substitution

The given equation is: $\left( \frac{9}{x} - \frac{9}{\sqrt{x}} + 2 \right) \left( \frac{2}{x} - \frac{7}{\sqrt{x}} + 3 \right) = 0$

Since we have $\sqrt{x}$ and $\frac{1}{x}$ terms, $x$ must be strictly positive, i.e., $x > 0$. This implies $\sqrt{x} > 0$.

Let $y = \frac{1}{\sqrt{x}}$. Then, $\frac{1}{x} = \left( \frac{1}{\sqrt{x}} \right)^2 = y^2$. Since $\sqrt{x} > 0$, we must also have $y > 0$.

Substituting $y$ into the equation, we get: $(9y^2 - 9y + 2)(2y^2 - 7y + 3) = 0$ This transforms the original equation into a product of two quadratic equations.

Step 2: Solve the First Quadratic Equation

The first quadratic equation is $9y^2 - 9y + 2 = 0$. We want to find the roots of this equation. We can factor this quadratic as follows: $9y^2 - 9y + 2 = 9y^2 - 6y - 3y + 2 = 3y(3y-2) - 1(3y-2) = (3y-1)(3y-2) = 0$ This gives us two possible solutions for $y$: $3y - 1 = 0 \implies y = \frac{1}{3}$ $3y - 2 = 0 \implies y = \frac{2}{3}$ Both solutions are positive, so they are valid.

Step 3: Solve the Second Quadratic Equation

The second quadratic equation is $2y^2 - 7y + 3 = 0$. We want to find the roots of this equation. We can factor this quadratic as follows: $2y^2 - 7y + 3 = 2y^2 - 6y - y + 3 = 2y(y-3) - 1(y-3) = (2y-1)(y-3) = 0$ This gives us two possible solutions for $y$: $2y - 1 = 0 \implies y = \frac{1}{2}$ $y - 3 = 0 \implies y = 3$ Both solutions are positive, so they are valid.

Step 4: Convert Back to x and Verify Solutions

We have four possible values for $y$: $\frac{1}{3}, \frac{2}{3}, \frac{1}{2}, 3$. We need to convert these back to $x$ using the substitution $y = \frac{1}{\sqrt{x}}$. This means $\sqrt{x} = \frac{1}{y}$, so $x = \frac{1}{y^2}$.

For $y = \frac{1}{3}$, $x = \frac{1}{(1/3)^2} = \frac{1}{1/9} = 9$. For $y = \frac{2}{3}$, $x = \frac{1}{(2/3)^2} = \frac{1}{4/9} = \frac{9}{4}$. For $y = \frac{1}{2}$, $x = \frac{1}{(1/2)^2} = \frac{1}{1/4} = 4$. For $y = 3$, $x = \frac{1}{3^2} = \frac{1}{9}$.

Since all four values of $x$ are positive, they are all valid solutions.

Step 5: Count the Solutions

We have found four distinct real values for $x$ that satisfy the original equation: $9, \frac{9}{4}, 4, \frac{1}{9}$.

Common Mistakes & Tips

• Forgetting the Domain: Always remember to check the domain of the original equation and the substituted variable. In this case, $x > 0$ and $y > 0$.
• Algebra Errors: Double-check your factoring and calculations to avoid mistakes.
• Not Converting Back: Remember to convert the solutions for the substituted variable back to the original variable to find the actual solutions.

Summary

We solved the given equation by using a substitution to transform it into a product of two quadratic equations. We solved each quadratic equation for the substituted variable, and then converted those solutions back to the original variable. We verified that all solutions were valid within the domain of the original equation. The total number of solutions is 4.

Final Answer The final answer is \boxed{4}, which corresponds to option (D).