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

Let $$\alpha - 1 = a^2$$ and $$\beta - 1 = b^2$$. Then $$\alpha = a^2+1$$ and $$\beta = b^2+1$$.

From $$x^2 - 70x + \lambda = 0$$:

• $$\alpha + \beta = 70 \implies a^2 + b^2 = 68$$

• $$\alpha\beta = \lambda \implies a^2b^2 + 69 = \lambda$$

The target expression simplifies to:

$$\frac{(a+b)(a^2b^2 + 104)}{|a^2-b^2|} = \frac{a^2b^2+104}{|a-b|} = \frac{a^2b^2+104}{\sqrt{68-2ab}}$$

 Let $$ab = t$$:

$$\frac{t^2+104}{\sqrt{68-2t}} = 60 \implies t = 16 \text{ works perfectly } \left(\frac{256+104}{\sqrt{36}} = \frac{360}{6} = 60\right)$$