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

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Question

(A) Given the cubic equation $x^3 - 2x^2 + 3x - 2 = 0$.By inspection,$x = 1$ is a root because $1 - 2 + 3 - 2 = 0$.Dividing the polynomial by $(x - 1)

Vedclass · Accepted Answer

$\frac{13}{4}$

Vedclass · Answer

(A) Given the cubic equation $x^3 - 2x^2 + 3x - 2 = 0$.By inspection,$x = 1$ is a root because $1 - 2 + 3 - 2 = 0$.Dividing the polynomial by $(x - 1)$,we get $(x - 1)(x^2 - x + 2) = 0$.Let $\alpha = 1$. Then $\beta$ and $\gamma$ 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 $\alpha = 1$: $S = \frac{\beta}{1 + \beta} + \frac{\gamma}{1 + \gamma} + \frac{\beta\gamma}{\beta + \gamma}$.Since $\beta + \gamma = 1$,the last term is $\frac{2}{1} = 2$.For the first two terms: $\frac{\beta(1 + \gamma) + \gamma(1 + \beta)}{(1 + \beta)(1 + \gamma)} = \frac{\beta + \beta\gamma + \gamma + \beta\gamma}{1 + \beta + \gamma + \beta\gamma} = \frac{(\beta + \gamma) + 2\beta\gamma}{1 + (\beta + \gamma) + \beta\gamma}$.Substituting the values: $\frac{1 + 2(2)}{1 + 1 + 2} = \frac{5}{4}$.Thus,$S = \frac{5}{4} + 2 = \frac{5 + 8}{4} = \frac{13}{4}$.

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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