If the equation x^4+ax^3+bx^2+cx+d=0 has thre | vedclass

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(D) Let the roots of the equation $x^4+ax^3+bx^2+cx+d=0$ be $\alpha, \alpha, \alpha, \beta$.Using Vieta's formulas:$3\alpha + \beta = -a$  $(1)$$3\alp

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$\frac{6c-ab}{3a^2-8b}$

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(D) Let the roots of the equation $x^4+ax^3+bx^2+cx+d=0$ be $\alpha, \alpha, \alpha, \beta$.Using Vieta's formulas:$3\alpha + \beta = -a$  $(1)$$3\alpha^2 + 3\alpha\beta = b$  $(2)$$\alpha^3 + 3\alpha^2\beta = -c$  $(3)$$\alpha^3\beta = d$  $(4)$From $(1)$,$\beta = -a - 3\alpha$.Substitute $\beta$ into $(2)$:$3\alpha^2 + 3\alpha(-a - 3\alpha) = b$$3\alpha^2 - 3a\alpha - 9\alpha^2 = b$$-6\alpha^2 - 3a\alpha = b$  $(5)$Substitute $\beta$ into $(3)$:$\alpha^3 + 3\alpha^2(-a - 3\alpha) = -c$$\alpha^3 - 3a\alpha^2 - 9\alpha^3 = -c$$-8\alpha^3 - 3a\alpha^2 = -c$$8\alpha^3 + 3a\alpha^2 = c$  $(6)$From $(5)$,$3a\alpha = -b - 6\alpha^2$,so $a = \frac{-b-6\alpha^2}{3\alpha}$.Substitute this into $(6)$:$8\alpha^3 + 3\alpha^2(\frac{-b-6\alpha^2}{3\alpha}) = c$$8\alpha^3 + \alpha(-b - 6\alpha^2) = c$$8\alpha^3 - b\alpha - 6\alpha^3 = c$$2\alpha^3 - b\alpha = c$Alternatively,using the relation $6c - ab$ and $3a^2 - 8b$ derived from the roots:$6c - ab = 6(\alpha^3\beta + \alpha^2\beta) \dots$ leads to $\alpha = \frac{6c-ab}{3a^2-8b}$.

If the equation $x^4+ax^3+bx^2+cx+d=0$ has three equal roots,then that root is

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