Let \(\alpha, \beta \in\) be roots of equation \(x^2-70 x+\lambda=0\), where \(\frac{\lambda}{2},…
Let \(\alpha, \beta \in\) be roots of equation \(x^2-70 x+\lambda=0\), where \(\frac{\lambda}{2}, \frac{\lambda}{3} \notin\). If \(\lambda\) assumes the minimum possible value, then \(\frac{(\sqrt{\alpha-1}+\sqrt{\beta-1})(\lambda+35)}{\boldsymbol{|\alpha-\beta|}}\) is equal to :
Solution
Given, $x^2 - 70x + \lambda = 0$
Sum of the roots will be $\alpha + \beta = 70$
Product of roots will be $\alpha \beta = \lambda$
$\Rightarrow \alpha (70 - \alpha) = \lambda$
$\Rightarrow \frac{d \lambda}{d \alpha} = 70 - 2\alpha$
Now, $\lambda$ is increasing for all $\alpha \leq 35$
And 2 and 3 does not divide $\lambda$ so taking $\alpha = 5$ we get, $\beta = 70 - \alpha = 65$
As for $\alpha = 1$, $\beta = 69$ & for $\alpha = 4$, $\beta = 66$ which is not possible as they are multiple of 2 & 3
Hence, $\alpha = 5$, $\beta = 65$, $\lambda = 325$
Now solving,
$\frac{\sqrt{\alpha - 1} + \sqrt{\beta - 1}}{|\alpha - \beta|} = \frac{\sqrt{5 - 1} + \sqrt{65 - 1}}{|5 - 65|} = \frac{2 + 8}{60}$ = 60
Alternative Solution:
The equation given is a quadratic equation, and the roots of this equation are $\alpha$ and $\beta$. By Vieta's formula, we know that the sum of the roots of a quadratic equation is equal to the opposite of the coefficient of $x$, divided by the coefficient of $x^2$. Thus, we have $\alpha + \beta = 70$.
The product of the roots of a quadratic equation is equal to the constant term, divided by the coefficient of $x^2$. In this case, the product of the roots, $\alpha \beta$, is equal to $\lambda$.
The condition given that $\frac{\lambda}{2}, $\frac{\lambda}{3}$ \notin \mathbb{Z}$ simply means that neither $\frac{\lambda}{2}$ nor $\frac{\lambda}{3}$ can be roots of the equation, i.e., $\lambda$ cannot be equal to $2\alpha$ or $2\beta$, nor can it be equal to $3\alpha$ or $3\beta$.
We want to find the minimum possible value of $\lambda$. Since $\lambda = \alpha \beta$, we want to find the minimum product of $\alpha$ and $\beta$, while satisfying the conditions.
Using AM-GM inequality (Arithmetic Mean - Geometric Mean inequality), we know that for any two positive numbers, the arithmetic mean is greater than or equal to the geometric mean, with equality if and only if the two numbers are equal.