The number of solutions of the equation $\left(\dfrac{9}{x}-\dfrac{9}{\sqrt{x}}+2\right)\!\left(\dfrac{2}{x}-\dfrac{7}{\sqrt{x}}+3\right)=0$ is:
Step-by-Step Solution
Key Concept: Substitute $\alpha = \dfrac{1}{\sqrt{x}}$ (valid since $x>0$). Each bracket becomes a quadratic in $\alpha$, and the product equals zero iff at least one bracket vanishes.
Put $\alpha = \dfrac{1}{\sqrt{x}}$, so $\dfrac{1}{x}=\alpha^{2}$. The equation reduces to
$$(9\alpha^{2}-9\alpha+2)(2\alpha^{2}-7\alpha+3)=0.$$
Factor each:
$$9\alpha^{2}-9\alpha+2 = (3\alpha-1)(3\alpha-2),\qquad 2\alpha^{2}-7\alpha+3=(2\alpha-1)(\alpha-3).$$
So $\alpha \in\left\{\dfrac{1}{3},\ \dfrac{2}{3},\ \dfrac{1}{2},\ 3\right\}$, giving
$$x=\frac{1}{\alpha^{2}}\in\{9,\ \tfrac{9}{4},\ 4,\ \tfrac{1}{9}\}.$$
All four are valid ($x>0$). Hence the number of solutions is $\boxed{4}$.
Correct Answer: 4