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

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Question

(B) 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

$\ln(1 + \frac{3}{5}x - \frac{1}{5}x^2)$

Vedclass · Answer

(B) 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 $S = (\alpha + \beta)x - \left( \frac{\alpha^2 + \beta^2}{2} \right)x^2 + \left( \frac{\alpha^3 + \beta^3}{3} \right)x^3 - \dots$We know the expansion $\ln(1 + t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \dots$Rewriting the series as: $S = (\alpha x - \frac{(\alpha x)^2}{2} + \frac{(\alpha x)^3}{3} - \dots) + (\beta x - \frac{(\beta x)^2}{2} + \frac{(\beta x)^3}{3} - \dots)$This simplifies to: $S = \ln(1 + \alpha x) + \ln(1 + \beta x) = \ln((1 + \alpha x)(1 + \beta x))$Expanding the product: $(1 + \alpha x)(1 + \beta x) = 1 + x(\alpha + \beta) + \alpha\beta x^2$Substituting the values $\alpha + \beta = \frac{3}{5}$ and $\alpha\beta = -\frac{1}{5}$,we get:$S = \ln(1 + \frac{3}{5}x - \frac{1}{5}x^2)$

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:

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