Let alpha and beta be the roots of 5x^2 - 3x | vedclass

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

Question

(D) Given the quadratic equation $5x^2 - 3x - 1 = 0$,the sum of roots is $\alpha + \beta = \frac{3}{5}$ and the product of roots is $\alpha\beta = -\f

Vedclass · Accepted Answer

none of these

Vedclass · Answer

(D) Given the quadratic equation $5x^2 - 3x - 1 = 0$,the sum of roots is $\alpha + \beta = \frac{3}{5}$ and the product of roots is $\alpha\beta = -\frac{1}{5}$.The given series is $S = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{\alpha^n + \beta^n}{n} x^n$.This can be split into two logarithmic series: $S = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(\alpha x)^n}{n} + \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(\beta x)^n}{n}$.Using the expansion $\ln(1+t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \dots$,we get $S = \ln(1 + \alpha x) + \ln(1 + \beta x)$.$S = \ln((1 + \alpha x)(1 + \beta x)) = \ln(1 + x(\alpha + \beta) + \alpha\beta x^2)$.Substituting the values: $S = \ln(1 + x(\frac{3}{5}) - \frac{1}{5}x^2) = \ln\left(\frac{5 + 3x - x^2}{5}\right)$.

Let $\alpha$ and $\beta$ be the roots of $5x^2 - 3x - 1 = 0$. Then,the expression $\left[ (\alpha + \beta)x - \left( \frac{\alpha^2 + \beta^2}{2} \right)x^2 + \left( \frac{\alpha^3 + \beta^3}{3} \right)x^3 - \dots \right]$ is equal to:

Similar Questions

$I. x^{2}-20 x+91=0$
$II. y^{2}-32 y+247=0$

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 $x^2 - 3x + 1 = 0$,then the equation whose roots are $\frac{1}{\alpha - 2}$ and $\frac{1}{\beta - 2}$ is

If the sum of the squares of the roots of the equation $x^2 - (\sin \alpha - 2)x - (1 + \sin \alpha) = 0$ is least,then $\alpha$ is

If $3p^2 = 5p + 2$ and $3q^2 = 5q + 2$,where $p \neq q$,then $pq$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module