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(D) Given the equation $x^4 - 100x^3 + 2x^2 + 4x + 10 = 0$ with roots $\alpha, \beta, \gamma, \delta$. To find the sum of the reciprocals $\frac{1}{\a

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$-\frac{2}{5}$

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(D) Given the equation $x^4 - 100x^3 + 2x^2 + 4x + 10 = 0$ with roots $\alpha, \beta, \gamma, \delta$. To find the sum of the reciprocals $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta}$,we can substitute $x = \frac{1}{y}$. Substituting into the equation: $(\frac{1}{y})^4 - 100(\frac{1}{y})^3 + 2(\frac{1}{y})^2 + 4(\frac{1}{y}) + 10 = 0$. Multiplying by $y^4$: $1 - 100y + 2y^2 + 4y^3 + 10y^4 = 0$,which is $10y^4 + 4y^3 + 2y^2 - 100y + 1 = 0$. The roots of this new equation are $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}, \frac{1}{\delta}$. The sum of the roots is given by $-\frac{	ext{coefficient of } y^3}{	ext{coefficient of } y^4} = -\frac{4}{10} = -\frac{2}{5}$.

If $\alpha, \beta, \gamma, \delta$ are the roots of $x^4 - 100x^3 + 2x^2 + 4x + 10 = 0$,then $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta}$ is equal to:

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