If alpha, beta, gamma are the roots of x^3-6x | vedclass

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Question

(A) Given the equation $x^3-6x^2+11x-6=0$. Factoring the cubic equation,we get $(x-1)(x-2)(x-3)=0$. Thus,the roots are $\alpha=1, \beta=2, \gamma=3$.

Vedclass · Accepted Answer

$x^3-28x^2+245x-650=0$

Vedclass · Answer

(A) Given the equation $x^3-6x^2+11x-6=0$. Factoring the cubic equation,we get $(x-1)(x-2)(x-3)=0$. Thus,the roots are $\alpha=1, \beta=2, \gamma=3$. Now,calculate the new roots: $\alpha' = \alpha^2+\beta^2 = 1^2+2^2 = 5$ $\beta' = \beta^2+\gamma^2 = 2^2+3^2 = 13$ $\gamma' = \gamma^2+\alpha^2 = 3^2+1^2 = 10$ The required equation is $x^3 - (\alpha'+\beta'+\gamma')x^2 + (\alpha'\beta'+\beta'\gamma'+\gamma'\alpha')x - \alpha'\beta'\gamma' = 0$. Sum of roots: $5+13+10 = 28$. Sum of roots taken two at a time: $(5 	imes 13) + (13 	imes 10) + (10 	imes 5) = 65 + 130 + 50 = 245$. Product of roots: $5 	imes 13 	imes 10 = 650$. Substituting these values,we get $x^3 - 28x^2 + 245x - 650 = 0$.

If $\alpha, \beta, \gamma$ are the roots of $x^3-6x^2+11x-6=0$,then the equation having the roots $\alpha^2+\beta^2, \beta^2+\gamma^2$ and $\gamma^2+\alpha^2$ is

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