Let (α, β ∈) be roots of equation (x^2-70 x+λ=0), where ((λ)/(2),...

Given,

${x}^{2} - 70x + \lambda = 0$

Sum of the roots will be $\alpha + \beta = 70$

Product of roots will be $\alpha\beta = \lambda$

$⇒ \alpha ( 70 - \alpha ) = \lambda$

$⇒ \dfrac{d\lambda}{d\alpha} = 70 - 2\alpha$

Now, $\lambda$ is increasing for all $\alpha \le 35$

And $2$ and $3$ does not divide $\lambda$ so taking  $\alpha = 5$ we get, $\beta = 70 - \alpha = 65$ 

$\text{As}\text{for}\alpha = 1, \beta = 69 \text{ and } \text{for}\alpha = 4, \beta = 66$ which is not possible as they are multiple of $2 \text{ and } 3$

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

Now solving,

$\dfrac{\left(\sqrt{\alpha - 1} + \sqrt{\beta - 1}\right)\left(\lambda + 35\right)}{| \alpha - \beta |}$

$= \dfrac{\left(\sqrt{5 - 1} + \sqrt{65 - 1}\right)\left(325 + 35\right)}{| 5 - 65 |}$

$= \dfrac{\left(2 + 8\right) \cdot 360}{60}$

$= 60$