The number of real roots of the equation sqrt | vedclass

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(B) Given the equation $\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6}$.Let $t = \sqrt{\frac{x}{1-x}}$. For $t$ to be defined and real,we must

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$2$

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(B) Given the equation $\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6}$.Let $t = \sqrt{\frac{x}{1-x}}$. For $t$ to be defined and real,we must have $\frac{x}{1-x} > 0$,which implies $x \in (0, 1)$.The equation becomes $t + \frac{1}{t} = \frac{13}{6}$.Multiplying by $6t$,we get $6t^2 - 13t + 6 = 0$.Factoring the quadratic: $(2t - 3)(3t - 2) = 0$,so $t = \frac{3}{2}$ or $t = \frac{2}{3}$.Case $1$: $\sqrt{\frac{x}{1-x}} = \frac{3}{2}$ $\Rightarrow \frac{x}{1-x} = \frac{9}{4}$ $\Rightarrow 4x = 9 - 9x$ $\Rightarrow 13x = 9$ $\Rightarrow x = \frac{9}{13}$.Case $2$: $\sqrt{\frac{x}{1-x}} = \frac{2}{3}$ $\Rightarrow \frac{x}{1-x} = \frac{4}{9}$ $\Rightarrow 9x = 4 - 4x$ $\Rightarrow 13x = 4$ $\Rightarrow x = \frac{4}{13}$.Both values $x = \frac{9}{13}$ and $x = \frac{4}{13}$ lie in the interval $(0, 1)$.Thus,there are $2$ real roots.

The number of real roots of the equation $\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{13}{6}$ is

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