If $\alpha$ satisfies the equation $\sqrt{\frac{x}{2 x+1}}+\sqrt{\frac{2 x+1}{x}}=2$, then the roots of the equation $\alpha^2 x^2+4 \alpha x+3=0$ are

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Aicharya · Accepted Answer

$\sqrt{\frac{x}{2 x+1}}+\sqrt{\frac{2 x+1}{x}}=2$ Let $\quad \sqrt{\frac{x}{2 x+1}}=y$ Then, $ y+\frac{1}{y}=2 $ $ \begin{aligned} & \Rightarrow \quad y^2-2 y+1=0 \ & \Rightarrow \quad(y-1)(y-1)=0 \ & \Rightarrow \quad(y-1)^2=0 \ & \Rightarrow \quad y-1=0 \ & \Rightarrow \quad y=1 \ & y=1 \ & \Rightarrow \quad \sqrt{\frac{x}{2 x+1}}=1 \ & \Rightarrow \quad \frac{x}{2 x+1}=1 \ & \Rightarrow \quad 2 x+1=x \ & \Rightarrow \quad x=-1 \ & \end{aligned} $ $\because \alpha$ satisfies the Eq. (i), therefore $\alpha=-1$ Now, $ \alpha^2 x^2+4 \alpha x+3=0 $ Put the value of $\alpha$ $ \begin{aligned} (-1)^2 x^2+4(-1) x+3 & =0 \ x^2-4 x+3 & =0 \ (x-1)(x-3) & =0 \ x & =1,3 \end{aligned} $ Hence, the roots of the equation $\alpha^2 x^2+4 \alpha x+3=0$ are 1,3

If $\alpha$ satisfies the equation $\sqrt{\frac{x}{2 x+1}}+\sqrt{\frac{2 x+1}{x}}=2$, then the roots of the…

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