Question

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

Answer 1

Analyze $\alpha$ and $\beta$:

Since $\alpha + \beta = 70$, assume possible integer values for $\alpha$ and $\beta$ that satisfy $\alpha \beta = \lambda$ and check divisibility conditions. A suitable pair is $\alpha = 5$ and $\beta = 65$, giving $\lambda = 5 \times 65 = 325$.

Verification:

Check that $\frac{325}{2} \notin \mathbb{N}$ and $\frac{325}{3} \notin \mathbb{N}$, satisfying the divisibility conditions.

Calculate the Required Expression:

Substitute $\alpha = 5$, $\beta = 65$, and $\lambda = 325$:

$\frac{\sqrt{\alpha - 1} + \sqrt{\beta - 1}(\lambda + 35)}{|\alpha - \beta|} = \frac{\sqrt{5 - 1} + \sqrt{65 - 1} \times (325 + 35)}{|5 - 65|}$

Simplifying this gives:

$= \frac{\sqrt{4 + 8 \times 360}}{60} = \frac{\sqrt{12 \times 360}}{60} = 60$