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

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(D) Let $S_1 = \alpha + \beta = -b/a$ and $S_2 = \alpha\beta = c/a$.Given $\alpha + \beta, \alpha^2 + \beta^2, \alpha^3 + \beta^3$ are in $G.P.$, so $

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$c\Delta = 0$

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(D) Let $S_1 = \alpha + \beta = -b/a$ and $S_2 = \alpha\beta = c/a$.Given $\alpha + \beta, \alpha^2 + \beta^2, \alpha^3 + \beta^3$ are in $G.P.$, so $(\alpha^2 + \beta^2)^2 = (\alpha + \beta)(\alpha^3 + \beta^3)$.We know $\alpha^2 + \beta^2 = S_1^2 - 2S_2$ and $\alpha^3 + \beta^3 = S_1^3 - 3S_1S_2$.Substituting these: $(S_1^2 - 2S_2)^2 = S_1(S_1^3 - 3S_1S_2) = S_1^4 - 3S_1^2S_2$.Expanding: $S_1^4 + 4S_2^2 - 4S_1^2S_2 = S_1^4 - 3S_1^2S_2$.Simplifying: $4S_2^2 - S_1^2S_2 = 0$, which implies $S_2(4S_2 - S_1^2) = 0$.Substituting $S_1 = -b/a$ and $S_2 = c/a$: $(c/a)(4(c/a) - (-b/a)^2) = 0$.$(c/a)((4ac - b^2)/a^2) = 0$.Since $\Delta = b^2 - 4ac$, we have $(c/a)(-\Delta/a^2) = 0$.Thus, $c\Delta = 0$.

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

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