Let alpha, beta in mathbb{N} be roots of the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the quadratic equation $x^2-70x+\lambda=0$ with roots $\alpha, \beta \in \mathbb{N}$.By Vieta's formulas,$\alpha+\beta=70$ and $\alpha\beta=

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(D) Given the quadratic equation $x^2-70x+\lambda=0$ with roots $\alpha, \beta \in \mathbb{N}$.By Vieta's formulas,$\alpha+\beta=70$ and $\alpha\beta=\lambda$.We are given that $\frac{\lambda}{2} 
otin \mathbb{N}$ and $\frac{\lambda}{3} 
otin \mathbb{N}$,which implies $\lambda$ is not divisible by $2$ or $3$. Thus,$\lambda$ is not divisible by $6$.Since $\lambda = \alpha(70-\alpha)$,we test values of $\alpha$ such that $\lambda$ is not divisible by $2$ or $3$.If $\alpha=1$,$\lambda=69$ (divisible by $3$,reject).If $\alpha=2$,$\lambda=136$ (divisible by $2$,reject).If $\alpha=3$,$\lambda=201$ (divisible by $3$,reject).If $\alpha=4$,$\lambda=264$ (divisible by $2$ and $3$,reject).If $\alpha=5$,$\lambda=5 	imes 65 = 325$. $325$ is not divisible by $2$ or $3$. This is the minimum value.For $\alpha=5, \beta=65, \lambda=325$,we calculate the expression:$|\alpha-\beta| = |5-65| = 60$.$\sqrt{\alpha-1} = \sqrt{4} = 2$.$\sqrt{\beta-1} = \sqrt{64} = 8$.Expression $= \frac{(2+8)(325+35)}{60} = \frac{10 	imes 360}{60} = 10 	imes 6 = 60$.

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 $\frac{(\sqrt{\alpha-1}+\sqrt{\beta-1})(\lambda+35)}{|\alpha-\beta|}$ is equal to :

Similar Questions

The number of solutions of the equation $x^2 + y^2 = a^2 + b^2 + c^2$,where $x, y, a, b, c$ are all prime numbers,is

If $ax + by = 1$,where $a, b, x$ and $y$ are integers,then which one of the following is not true?

The minimum value of $\frac{9 \cdot 3^{2x} + 6 \cdot 3^x + 4}{9 \cdot 3^{2x} - 6 \cdot 3^x + 4}$ is

If $\alpha, \beta$ are the roots of the equation $x^2 - px + q = 0$,then the quadratic equation whose roots are $(\alpha^2 - \beta^2)(\alpha^3 - \beta^3)$ and $\alpha^3\beta^2 + \alpha^2\beta^3$ is (where $S = p[p^4 - 5p^2q + 5q^2]$ and $P = p^2q^2(p^4 - 5p^2q + 4q^2)$).

For $x \in R$, the least value of $\frac{x^2-6x+5}{x^2+2x+1}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module