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(C) Given equations are:$x^3-2x-25\lambda=0 \quad (1)$$3x^3-8x-\frac{175}{3}\lambda=0 \quad (2)$Since $\alpha$ is a common root,we have:$25\lambda = \

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$\frac{3}{5\sqrt{5}}$

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(C) Given equations are:$x^3-2x-25\lambda=0 \quad (1)$$3x^3-8x-\frac{175}{3}\lambda=0 \quad (2)$Since $\alpha$ is a common root,we have:$25\lambda = \alpha^3-2\alpha \Rightarrow \lambda = \frac{\alpha^3-2\alpha}{25}$$\frac{175}{3}\lambda = 3\alpha^3-8\alpha$ $\Rightarrow \lambda = \frac{3(3\alpha^3-8\alpha)}{175} = \frac{9\alpha^3-24\alpha}{175}$Equating the two expressions for $\lambda$:$\frac{\alpha^3-2\alpha}{25} = \frac{9\alpha^3-24\alpha}{175}$Multiply both sides by $175$:$7(\alpha^3-2\alpha) = 9\alpha^3-24\alpha$$7\alpha^3-14\alpha = 9\alpha^3-24\alpha$$2\alpha^3-10\alpha = 0$$2\alpha(\alpha^2-5) = 0$Since $\lambda > 0$,we check $\alpha^2=5$,i.e.,$\alpha = \pm\sqrt{5}$.If $\alpha = \sqrt{5}$,then $\lambda = \frac{(\sqrt{5})^3-2\sqrt{5}}{25} = \frac{5\sqrt{5}-2\sqrt{5}}{25} = \frac{3\sqrt{5}}{25} = \frac{3}{5\sqrt{5}}$.If $\alpha = -\sqrt{5}$,then $\lambda = \frac{-5\sqrt{5}+2\sqrt{5}}{25} = \frac{-3\sqrt{5}}{25} Thus,$\lambda = \frac{3}{5\sqrt{5}}$.

Let $\alpha$ be a common root of the equations $x^3-2x-25\lambda=0$ and $3x^3-8x-\frac{175}{3}\lambda=0$,where $\lambda > 0$. Then $\lambda=$

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