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(D) Given the equation $x^4 - 4x^3 + ax^2 + bx + 1 = 0$ has four real roots $\alpha, \beta, \gamma, \delta$.By Vieta's formulas,we have:$\sum \alpha =

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$6, -4$

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(D) Given the equation $x^4 - 4x^3 + ax^2 + bx + 1 = 0$ has four real roots $\alpha, \beta, \gamma, \delta$.By Vieta's formulas,we have:$\sum \alpha = 4$$\sum \alpha \beta = a$$\sum \alpha \beta \gamma = -b$$\alpha \beta \gamma \delta = 1$For real positive roots,the Arithmetic Mean $(AM)$ is greater than or equal to the Geometric Mean $(GM)$:$\frac{\alpha + \beta + \gamma + \delta}{4} \geq (\alpha \beta \gamma \delta)^{1/4}$Substituting the values: $\frac{4}{4} \geq (1)^{1/4} \implies 1 \geq 1$.Since the $AM$ equals the $GM$,all roots must be equal: $\alpha = \beta = \gamma = \delta = 1$.Now,calculate $a$ and $b$:$a = \sum \alpha \beta = \binom{4}{2} 	imes (1 	imes 1) = 6 	imes 1 = 6$.$-b = \sum \alpha \beta \gamma = \binom{4}{3} 	imes (1 	imes 1 	imes 1) = 4 	imes 1 = 4 \implies b = -4$.Thus,$a = 6$ and $b = -4$.

If the equation $x^4 - 4x^3 + ax^2 + bx + 1 = 0$ has four real roots $\alpha, \beta, \gamma, \delta$,then the values of $a$ and $b$ are:

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