If alpha, beta are the roots of the equation | vedclass

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(A) Given that $\alpha$ and $\beta$ are the roots of $x^2 - px + q = 0$,we have $\alpha + \beta = p$ and $\alpha \beta = q$.Now,consider the expressio

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$(\alpha + \beta)x - \frac{\alpha^2 + \beta^2}{2}x^2 + \frac{\alpha^3 + \beta^3}{3}x^3 - \dots \infty$

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(A) Given that $\alpha$ and $\beta$ are the roots of $x^2 - px + q = 0$,we have $\alpha + \beta = p$ and $\alpha \beta = q$.Now,consider the expression $\log_e(1 + px + qx^2)$.Substituting $p = \alpha + \beta$ and $q = \alpha \beta$,we get:$\log_e(1 + (\alpha + \beta)x + \alpha \beta x^2) = \log_e((1 + \alpha x)(1 + \beta x))$.Using the property $\log(ab) = \log a + \log b$,we have:$\log_e(1 + \alpha x) + \log_e(1 + \beta x)$.Using the logarithmic series expansion $\log_e(1 + t) = t - \frac{t^2}{2} + \frac{t^3}{3} - \dots$,we get:$\left( \alpha x - \frac{(\alpha x)^2}{2} + \frac{(\alpha x)^3}{3} - \dots \right) + \left( \beta x - \frac{(\beta x)^2}{2} + \frac{(\beta x)^3}{3} - \dots \right)$.Grouping the terms by powers of $x$:$= (\alpha + \beta)x - \frac{\alpha^2 + \beta^2}{2}x^2 + \frac{\alpha^3 + \beta^3}{3}x^3 - \dots \infty$.

If $\alpha, \beta$ are the roots of the equation $x^2 - px + q = 0$,then $\log_e(1 + px + qx^2) = $

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