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

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$\frac{-2}{5}$

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(D) Given the polynomial 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 $x = \frac{1}{y}$ into the original equation.Substituting $x = \frac{1}{y}$ gives: $(\frac{1}{y})^4 - 100(\frac{1}{y})^3 + 2(\frac{1}{y})^2 + 4(\frac{1}{y}) + 10 = 0$.Multiplying the entire equation by $y^4$,we get: $1 - 100y + 2y^2 + 4y^3 + 10y^4 = 0$,which rearranges to $10y^4 + 4y^3 + 2y^2 - 100y + 1 = 0$.The roots of this new equation are $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}, \frac{1}{\delta}$.The sum of the roots of a polynomial $a_n x^n + a_{n-1} x^{n-1} + \dots + a_0 = 0$ is given by $-\frac{a_{n-1}}{a_n}$.Applying this to our new equation,the sum is $-\frac{4}{10} = -\frac{2}{5}$.

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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