Question

\(\text{The number of solutions of the equation}\)\(\left(\frac{9}{x}-\frac{9}{\sqrt{x}}+2\right)\left(\frac{2}{x}-\frac{7}{\sqrt{x}}+3\right)=0\mathrm \; {is:}\)

• A. 3
• B. 1
• C. 4
• D. 2

Hint:

When dealing with square roots in equations, try substitution to simplify the expression and reduce it to a quadratic form.

Answer 1

The Correct Option is C

We are tasked with solving the given equation and determining the number of solutions for \( x \). Let us proceed step by step:

1. Problem Setup:
We are given the substitution:

\( \frac{1}{\sqrt{x}} = \alpha, \quad x > 0 \)

The equation to solve is:

\( (9\alpha^2 - 9\alpha + 2)(2\alpha^2 - 7\alpha + 3) = 0 \)

2. Factoring the Quartic Equation:
The quartic equation can be factored as:

\( (3\alpha - 2)(3\alpha - 1)(\alpha - 3)(2\alpha - 1) = 0 \)

3. Solving for \( \alpha \):
Set each factor equal to zero to find the values of \( \alpha \):

\( 3\alpha - 2 = 0 \implies \alpha = \frac{2}{3} \)

\( 3\alpha - 1 = 0 \implies \alpha = \frac{1}{3} \)

\( \alpha - 3 = 0 \implies \alpha = 3 \)

\( 2\alpha - 1 = 0 \implies \alpha = \frac{1}{2} \)

Thus, the solutions for \( \alpha \) are:

\( \alpha = \frac{1}{3}, \frac{1}{2}, \frac{2}{3}, 3 \)

4. Substituting Back to Find \( x \):
Recall that \( \alpha = \frac{1}{\sqrt{x}} \). Solving for \( x \), we get:

\( x = \frac{1}{\alpha^2} \)

Substitute each value of \( \alpha \):

\( \alpha = \frac{1}{3} \implies x = \frac{1}{\left(\frac{1}{3}\right)^2} = 9 \)

\( \alpha = \frac{1}{2} \implies x = \frac{1}{\left(\frac{1}{2}\right)^2} = 4 \)

\( \alpha = \frac{2}{3} \implies x = \frac{1}{\left(\frac{2}{3}\right)^2} = \frac{9}{4} \)

\( \alpha = 3 \implies x = \frac{1}{3^2} = \frac{1}{9} \)

Thus, the corresponding values of \( x \) are:

\( x = 9, 4, \frac{9}{4}, \frac{1}{9} \)

5. Counting the Number of Solutions:
There are 4 distinct values of \( x \). Therefore, the total number of solutions is:

\( \text{Number of solutions} = 4 \)

Final Answer:
The number of solutions is \( \boxed{4} \).