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(A) Given the cubic equation $x^3-4x^2-9x+36=0$ with roots $\alpha, \beta, \gamma$.From the relations between roots and coefficients:$1) \alpha+\beta+

Vedclass · Accepted Answer

$75$

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(A) Given the cubic equation $x^3-4x^2-9x+36=0$ with roots $\alpha, \beta, \gamma$.From the relations between roots and coefficients:$1) \alpha+\beta+\gamma = -(-4)/1 = 4$$2) \alpha\beta+\beta\gamma+\gamma\alpha = -9/1 = -9$$3) \alpha\beta\gamma = -36/1 = -36$Given $\alpha+\beta=0$,substituting this into $(1)$ gives $0+\gamma=4$,so $\gamma=4$.Substituting $\gamma=4$ into $(2)$: $\alpha\beta + 4(\alpha+\beta) = -9$. Since $\alpha+\beta=0$,we get $\alpha\beta = -9$.Substituting $\gamma=4$ into $(3)$: $\alpha\beta(4) = -36$,which gives $\alpha\beta = -9$. This is consistent.Since $\alpha+\beta=0$,$\beta=-\alpha$. Then $\alpha(-\alpha) = -9 \Rightarrow \alpha^2 = 9$,so $\alpha = 3$ or $\alpha = -3$.If $\alpha=3$,then $\beta=-3$. If $\alpha=-3$,then $\beta=3$.In either case,$\alpha^2=9$ and $\beta^2=9$.Now,calculate $\alpha^2+2\beta^2+3\gamma^2 = 9 + 2(9) + 3(4^2) = 9 + 18 + 3(16) = 27 + 48 = 75$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-4x^2-9x+36=0$ such that $\alpha+\beta=0$,then $\alpha^2+2\beta^2+3\gamma^2=$

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