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(D) Given the equation $\sqrt{\frac{5x}{x-2}} + \sqrt{\frac{x-2}{5x}} = \frac{29}{10}$.Let $y = \sqrt{\frac{5x}{x-2}}$. Then the equation becomes $y +

Vedclass · Accepted Answer

$6$

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(D) Given the equation $\sqrt{\frac{5x}{x-2}} + \sqrt{\frac{x-2}{5x}} = \frac{29}{10}$.Let $y = \sqrt{\frac{5x}{x-2}}$. Then the equation becomes $y + \frac{1}{y} = \frac{29}{10}$.Multiplying by $10y$,we get $10y^2 - 29y + 10 = 0$.Factoring the quadratic: $10y^2 - 25y - 4y + 10 = 0 \Rightarrow 5y(2y - 5) - 2(2y - 5) = 0$.So,$(5y - 2)(2y - 5) = 0$,which gives $y = \frac{2}{5}$ or $y = \frac{5}{2}$.Case $1$: $\sqrt{\frac{5x}{x-2}} = \frac{2}{5}$ $\Rightarrow \frac{5x}{x-2} = \frac{4}{25}$ $\Rightarrow 125x = 4x - 8$ $\Rightarrow 121x = -8$ $\Rightarrow x = -\frac{8}{121}$.Case $2$: $\sqrt{\frac{5x}{x-2}} = \frac{5}{2}$ $\Rightarrow \frac{5x}{x-2} = \frac{25}{4}$ $\Rightarrow \frac{x}{x-2} = \frac{5}{4}$ $\Rightarrow 4x = 5x - 10$ $\Rightarrow x = 10$.Since $\alpha > \beta$,we have $\alpha = 10$ and $\beta = -\frac{8}{121}$.Now,calculate $\sqrt{\alpha^2 - 11^4 \beta^2} = \sqrt{10^2 - 11^4 \left(-\frac{8}{121}\right)^2} = \sqrt{100 - 11^4 \cdot \frac{64}{11^4}} = \sqrt{100 - 64} = \sqrt{36} = 6$.

If $\alpha$ and $\beta$ are the real roots of the equation $\sqrt{\frac{5x}{x-2}} + \sqrt{\frac{x-2}{5x}} = \frac{29}{10}$ and $\alpha > \beta$,then $\sqrt{\alpha^2 - 11^4 \beta^2} = $

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