If alpha satisfies the equation sqrt{frac{x}{ | vedclass

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(A) Given equation: $\sqrt{\frac{x}{2x+1}} + \sqrt{\frac{2x+1}{x}} = 2$ Let $y = \sqrt{\frac{x}{2x+1}}$. Then the equation becomes $y + \frac{1}{y} =

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$1, 3$

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(A) Given equation: $\sqrt{\frac{x}{2x+1}} + \sqrt{\frac{2x+1}{x}} = 2$ Let $y = \sqrt{\frac{x}{2x+1}}$. Then the equation becomes $y + \frac{1}{y} = 2$. Multiplying by $y$,we get $y^2 - 2y + 1 = 0$,which is $(y-1)^2 = 0$. Thus,$y = 1$. Substituting back: $\sqrt{\frac{x}{2x+1}} = 1 \implies \frac{x}{2x+1} = 1 \implies x = 2x + 1 \implies x = -1$. Since $\alpha$ satisfies this equation,$\alpha = -1$. Now,substitute $\alpha = -1$ into the quadratic equation $\alpha^2 x^2 + 4\alpha x + 3 = 0$: $(-1)^2 x^2 + 4(-1)x + 3 = 0 \implies x^2 - 4x + 3 = 0 \implies (x-1)(x-3) = 0$. Therefore,the roots are $x = 1, 3$.

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

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