If alpha, beta are the roots of x^2+ax+2=0 an | vedclass

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

Question

(A) Given $x^2+ax+2=0$,we have $\alpha+\beta = -a$ and $\alpha\beta = 2$. For $x^2-bx+c=0$,the roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$,so $

Vedclass · Accepted Answer

$\frac{9}{4}(9-a^2)$

Vedclass · Answer

(A) Given $x^2+ax+2=0$,we have $\alpha+\beta = -a$ and $\alpha\beta = 2$. For $x^2-bx+c=0$,the roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$,so $\frac{1}{\alpha}+\frac{1}{\beta} = b$ and $\frac{1}{\alpha\beta} = c = \frac{1}{2}$. Now,consider the expression: $E = \left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right)$ $E = \left(\alpha\beta + 1 + 1 + \frac{1}{\alpha\beta}\right) \left(\alpha\beta - 1 - 1 + \frac{1}{\alpha\beta}\right)$ $E = \left(\alpha\beta + \frac{1}{\alpha\beta} + 2\right) \left(\alpha\beta + \frac{1}{\alpha\beta} - 2\right)$ $E = \left(\alpha\beta + \frac{1}{\alpha\beta}\right)^2 - 4$ Substituting $\alpha\beta = 2$: $E = \left(2 + \frac{1}{2}\right)^2 - 4 = \left(\frac{5}{2}\right)^2 - 4 = \frac{25}{4} - 4 = \frac{9}{4}$. Wait,let us re-evaluate the expression expansion: $E = \left(\alpha\beta + 1 + 1 + \frac{1}{\alpha\beta}\right) \left(\alpha\beta - \frac{\alpha}{\beta} - \frac{\beta}{\alpha} + \frac{1}{\alpha\beta}\right)$ $E = \left(2 + 2 + \frac{1}{2}\right) \left(2 + \frac{1}{2} - \frac{\alpha^2+\beta^2}{\alpha\beta}\right) = \frac{9}{2} \left(\frac{5}{2} - \frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta}\right)$ $E = \frac{9}{2} \left(\frac{5}{2} - \frac{a^2-4}{2}\right) = \frac{9}{2} \left(\frac{5-a^2+4}{2}\right) = \frac{9}{4}(9-a^2)$.

If $\alpha, \beta$ are the roots of $x^2+ax+2=0$ and $\frac{1}{\alpha}, \frac{1}{\beta}$ are the roots of $x^2-bx+c=0$,then $\left(\alpha+\frac{1}{\beta}\right)\left(\beta+\frac{1}{\alpha}\right)\left(\alpha-\frac{1}{\alpha}\right)\left(\beta-\frac{1}{\beta}\right) = $

Similar Questions

If the harmonic mean between the roots of $(5+\sqrt{2}) x^2-b x+(8+2 \sqrt{5})=0$ is $4$,then the value of $b$ is

If $\frac{\alpha}{\alpha+1}$ and $\frac{\beta}{\beta+1}$ are the roots of the quadratic equation $x^2+7x+3=0$,then the equation having roots $\alpha$ and $\beta$ is

If $\alpha$ and $\beta$ are the roots of $x^2-10x-8=0$ with $\alpha > \beta$,and $a_n = \alpha^n - \beta^n$ for $n \in N$,then the value of $\frac{a_{10}-8a_8}{5a_9}$ is:

If $\alpha, \beta$ are the roots of $(x - a)(x - b) = c$,where $c \neq 0$,then the roots of $(x - \alpha)(x - \beta) + c = 0$ are

Let two numbers have an arithmetic mean of $9$ and a geometric mean of $4$. Then these numbers are the roots of the quadratic equation:

Vedclass Test Series

Exam Paper Generator

Online Exam Module