The number of solutions of the equation left( | vedclass

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(B) Let $\frac{1}{\sqrt{x}} = \alpha$,where $x > 0$. Substituting this into the equation,we get: $(9\alpha^2 - 9\alpha + 2)(2\alpha^2 - 7\alpha + 3) =

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Let $\frac{1}{\sqrt{x}} = \alpha$,where $x > 0$. Substituting this into the equation,we get: $(9\alpha^2 - 9\alpha + 2)(2\alpha^2 - 7\alpha + 3) = 0$. Factoring the quadratic expressions: $(3\alpha - 1)(3\alpha - 2)(2\alpha - 1)(\alpha - 3) = 0$. This gives the roots for $\alpha$ as $\alpha = \frac{1}{3}, \frac{2}{3}, \frac{1}{2}, 3$. Since $\alpha = \frac{1}{\sqrt{x}}$,we have $\sqrt{x} = \frac{1}{\alpha}$,so $x = \frac{1}{\alpha^2}$. Calculating $x$ for each value of $\alpha$: For $\alpha = \frac{1}{3}, x = 9$. For $\alpha = \frac{2}{3}, x = \frac{9}{4}$. For $\alpha = \frac{1}{2}, x = 4$. For $\alpha = 3, x = \frac{1}{9}$. All these values of $x$ are positive and satisfy the condition $x > 0$. Thus,there are $4$ solutions.

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

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