"Let $\alpha, \beta \in \mathbf{N}$ be roots of the equation $x^2-70 x+\lambda=0$, where $\frac{\lambda}{2}, \frac{\lambda}{3} \notin \mathbf{N}$. If $\lambda$ assumes the minimum possible value, then $\frac{(\sqrt{\alpha-1}+\sqrt{\beta-1})(\lambda+35)}{|\alpha-\beta|}$ is equal to :"

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$$\begin{aligned} & x^2-70 x+\lambda=0 \\ & \alpha+\beta=70 \\ & \alpha \beta=\lambda \\ & \therefore \alpha(70-\alpha)=\lambda \end{aligned}$$

Since, 2 and 3 does not divide $\lambda$

$\therefore \alpha=5, \beta=65, \lambda=325$

By putting value of $\alpha, \beta, \lambda$ we get the required value $60$.

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Let $\alpha, \beta \in \mathbf{N}$ be roots of the equation $x^2-70 x+\lambda=0$, where $\frac{\lambda}{2}, \frac{\lambda}{3} \notin \mathbf{N}$. If $\lambda$ assumes the minimum possible value, then $\frac{(\sqrt{\alpha-1}+\sqrt{\beta-1})(\lambda+35)}{|\alpha-\beta|}$ is equal to :

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Let $\alpha, \beta \in \mathbf{N}$ be roots of the equation $x^2-70 x+\lambda=0$, where $\frac{\lambda}{2}, \frac{\lambda}{3} \notin \mathbf{N}$. If $\lambda$ assumes the minimum possible value, then $\frac{(\sqrt{\alpha-1}+\sqrt{\beta-1})(\lambda+35)}{|\alpha-\beta|}$ is equal to :

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