If alpha, beta, gamma are roots of x^3 - 2x^2 | vedclass

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Question

(A) Given the equation $x^3 - 2x^2 + 3x - 2 = 0$. By inspection,$x = 1$ is a root. Dividing by $(x - 1)$,we get $(x - 1)(x^2 - x + 2) = 0$. Thus,$\alp

Vedclass · Accepted Answer

$\frac{13}{4}$

Vedclass · Answer

(A) Given the equation $x^3 - 2x^2 + 3x - 2 = 0$. By inspection,$x = 1$ is a root. Dividing by $(x - 1)$,we get $(x - 1)(x^2 - x + 2) = 0$. Thus,$\alpha = 1$ and $\beta, \gamma$ are roots of $x^2 - x + 2 = 0$. From the quadratic equation,$\beta + \gamma = 1$ and $\beta \gamma = 2$. We need to evaluate $S = \frac{\alpha \beta}{\alpha + \beta} + \frac{\alpha \gamma}{\alpha + \gamma} + \frac{\beta \gamma}{\beta + \gamma}$. Substituting $\alpha = 1$: $S = \frac{\beta}{1 + \beta} + \frac{\gamma}{1 + \gamma} + \frac{2}{1} = \frac{\beta(1 + \gamma) + \gamma(1 + \beta)}{(1 + \beta)(1 + \gamma)} + 2$. $S = \frac{\beta + \beta \gamma + \gamma + \beta \gamma}{1 + (\beta + \gamma) + \beta \gamma} + 2 = \frac{(\beta + \gamma) + 2(\beta \gamma)}{1 + (\beta + \gamma) + \beta \gamma} + 2$. Substituting values: $S = \frac{1 + 2(2)}{1 + 1 + 2} + 2 = \frac{5}{4} + 2 = \frac{13}{4}$.

If $\alpha, \beta, \gamma$ are roots of $x^3 - 2x^2 + 3x - 2 = 0$,then the value of $\left( \frac{\alpha \beta}{\alpha + \beta} + \frac{\alpha \gamma}{\alpha + \gamma} + \frac{\beta \gamma}{\beta + \gamma} \right)$ is

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