If alpha and beta are the roots of the equati | vedclass

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

Question

(D) Given,$\alpha$ and $\beta$ are the roots of $2x^2 + 6x + k = 0$. From the relation between roots and coefficients,$\alpha + \beta = -\frac{6}{2} =

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(D) Given,$\alpha$ and $\beta$ are the roots of $2x^2 + 6x + k = 0$. From the relation between roots and coefficients,$\alpha + \beta = -\frac{6}{2} = -3$ and $\alpha\beta = \frac{k}{2}$. Now,consider the expression $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta}$. Substituting the values,we get $\frac{(-3)^2 - 2(k/2)}{k/2} = \frac{9 - k}{k/2} = \frac{18 - 2k}{k} = \frac{18}{k} - 2$. Since $k As $k$ becomes more negative (i.e.,$k 	o -\infty$),$\frac{18}{k} 	o 0$,so $f(k) 	o -2$. For $k Specifically,for $k However,for $-18 \le k The maximum value of the greatest integer function $\left[\frac{18}{k} - 2ight]$ for $k < 0$ is $-2$.

If $\alpha$ and $\beta$ are the roots of the equation $2x^2 + 6x + k = 0$,then the maximum value of $\left[\frac{\alpha}{\beta} + \frac{\beta}{\alpha}\right]$ when $k < 0$ is (where $[\cdot]$ denotes the greatest integer function)

Similar Questions

For the equation $2x^2 - 2(m^2 + 1)x + m^4 + m^2 + 1 = 0$,if $\alpha$ and $\beta$ are the roots,then $\alpha^2 + \beta^2 = \dots$

If the difference of the roots of $x^2 - px + 8 = 0$ is $2$,then the value of $p$ is

If $\alpha$ and $\beta$ are roots of $ax^2 + 2bx + c = 0$,then $\sqrt{\frac{\alpha}{\beta}} + \sqrt{\frac{\beta}{\alpha}}$ is equal to

The quadratic equation in $t$,such that the $A.M.$ of its roots is $A$ and the $G.M.$ of its roots is $G$,is

If $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 + bx - c = 0$,then the equation whose roots are $b$ and $c$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module