If alpha and beta are the roots of ax^2 + bx | vedclass

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(D) Given that $\alpha + \beta$,$\alpha^2 + \beta^2$,and $\alpha^3 + \beta^3$ are in $G.P.$Therefore,$(\alpha^2 + \beta^2)^2 = (\alpha + \beta)(\alpha

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$c\Delta = 0$

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(D) Given that $\alpha + \beta$,$\alpha^2 + \beta^2$,and $\alpha^3 + \beta^3$ are in $G.P.$Therefore,$(\alpha^2 + \beta^2)^2 = (\alpha + \beta)(\alpha^3 + \beta^3)$.Using the relations $\alpha + \beta = -b/a$ and $\alpha\beta = c/a$,we have:$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = b^2/a^2 - 2c/a = (b^2 - 2ac)/a^2$.$\alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta) = (-b/a)(b^2/a^2 - 3c/a) = (-b^3 + 3abc)/a^3$.Substituting these into the $G.P.$ condition:$((b^2 - 2ac)/a^2)^2 = (-b/a)((-b^3 + 3abc)/a^3)$.$(b^2 - 2ac)^2 / a^4 = (b^4 - 3abc^2)/a^4$ (Correction: $(b^4 - 3ab^2c)/a^4$ is incorrect,let's simplify: $(b^2 - 2ac)^2 = b^4 - 3ab^2c$ is wrong. Correct expansion: $b^4 + 4a^2c^2 - 4ab^2c = b^4 - 3ab^2c$).$4a^2c^2 - ab^2c = 0$.$ac(4ac - b^2) = 0$.Since $a 
eq 0$,we have $c(4ac - b^2) = 0$,which implies $c\Delta = 0$.

If $\alpha$ and $\beta$ are the roots of $ax^2 + bx + c = 0$ and $\alpha + \beta$,$\alpha^2 + \beta^2$,and $\alpha^3 + \beta^3$ are in $G.P.$,where $\Delta = b^2 - 4ac$,then:

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