If the roots of $\sqrt{\frac{1-y}{y}}+\sqrt{\frac{y}{1-y}}=\frac{5}{2}$ are $\alpha$ and $\beta(\beta\gt\alpha)$ and the equation $(\alpha+\beta) x^4-25 \alpha \beta x^2+(\gamma+\beta-\alpha)=0$ has real roots, then a possible value of $\gamma$ is

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Aicharya · Accepted Answer

$\sqrt{\frac{1-y}{y}}+\sqrt{\frac{y}{1-y}}=\frac{5}{2}$ $\begin{aligned} & \frac{1}{\sqrt{y(1-y)}}=\frac{5}{2} \Rightarrow y(1-y)=\frac{4}{25} \ & 25 y(1-y)=4 \Rightarrow 25 y^2-25 y+4=0 \ & \alpha+\beta=1 \ & \alpha \beta=\frac{4}{25} \ & \beta-\alpha=\sqrt{1-\frac{16}{25}}=\frac{3}{5} \ & x^4-4 x^2+\left(\frac{3}{5}+\gammaight)=0 \ & x^2=\frac{4 \pm \sqrt{16-4\left(\frac{3}{5}+\gammaight)}}{2}=\frac{4 \pm 2 \sqrt{\frac{17}{5}-\gamma}}{2}\end{aligned}$ $\begin{aligned} & 	herefore 2+\sqrt{\frac{17}{5}-\gamma}\gt0 \forall \gamma \in \mathrm{R} \ & 2-\sqrt{\frac{17}{5}-\gamma\gt0} \Rightarrow 2\gt\sqrt{\frac{17}{5}-\gamma} \ & 4\gt\frac{17}{5}-\gamma \Rightarrow \gamma\gt-\frac{3}{5} \Rightarrow \frac{17}{5}-\gamma\gt0 \Rightarrow \gamma \lt \frac{17}{5} \ & \gamma \in\left(\frac{-3}{5}, \frac{17}{5}ight)\end{aligned}$

If the roots of $\sqrt{\frac{1-y}{y}}+\sqrt{\frac{y}{1-y}}=\frac{5}{2}$ are $\alpha$ and…

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