If the roots of the equation sqrt{frac{x}{1-x | vedclass

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(B) Let $y = \sqrt{\frac{x}{1-x}}$. Then the equation becomes $y + \frac{1}{y} = \frac{5}{2}$.Multiplying by $2y$,we get $2y^2 - 5y + 2 = 0$,which fac

Vedclass · Accepted Answer

$\frac{104}{25}$

Vedclass · Answer

(B) Let $y = \sqrt{\frac{x}{1-x}}$. Then the equation becomes $y + \frac{1}{y} = \frac{5}{2}$.Multiplying by $2y$,we get $2y^2 - 5y + 2 = 0$,which factors as $(2y-1)(y-2) = 0$.So $y = 2$ or $y = \frac{1}{2}$.If $y = 2$,then $\frac{x}{1-x} = 4$ $\Rightarrow x = 4 - 4x$ $\Rightarrow 5x = 4$ $\Rightarrow x = \frac{4}{5}$.If $y = \frac{1}{2}$,then $\frac{x}{1-x} = \frac{1}{4}$ $\Rightarrow 4x = 1 - x$ $\Rightarrow 5x = 1$ $\Rightarrow x = \frac{1}{5}$.Given $p > q$,we have $p = \frac{4}{5}$ and $q = \frac{1}{5}$.Then $p+q = 1$,$pq = \frac{4}{25}$,and $\frac{p}{q} = 4$.The second equation is $1x^4 - \frac{4}{25}x^2 + 4 = 0$,or $25x^4 - 4x^2 + 100 = 0$.For a polynomial $ax^4 + bx^3 + cx^2 + dx + e = 0$,the sum of roots $\Sigma \alpha = -\frac{b}{a} = 0$.The sum of roots taken two at a time $\Sigma \alpha \beta = \frac{c}{a} = \frac{-4}{25}$.The product of roots $\alpha \beta \gamma \delta = \frac{e}{a} = \frac{100}{25} = 4$.Therefore,$(\Sigma \alpha)^2 - \Sigma \alpha \beta + \alpha \beta \gamma \delta = (0)^2 - (-\frac{4}{25}) + 4 = \frac{4}{25} + 4 = \frac{4 + 100}{25} = \frac{104}{25}$.

If the roots of the equation $\sqrt{\frac{x}{1-x}}+\sqrt{\frac{1-x}{x}}=\frac{5}{2}$ are $p$ and $q$ $(p > q)$ and the roots of the equation $(p+q)x^4 - pqx^2 + \frac{p}{q} = 0$ are $\alpha, \beta, \gamma, \delta$,then $(\Sigma \alpha)^2 - \Sigma \alpha \beta + \alpha \beta \gamma \delta = $

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