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(D) Let the roots be $\alpha, \beta, \gamma$. From the given equation $x^3-12x^2+kx-18=0$,we have the relations: $\alpha+\beta+\gamma = 12$ $\alpha\be

Vedclass · Accepted Answer

$57$

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(D) Let the roots be $\alpha, \beta, \gamma$. From the given equation $x^3-12x^2+kx-18=0$,we have the relations: $\alpha+\beta+\gamma = 12$ $\alpha\beta+\beta\gamma+\gamma\alpha = k$ $\alpha\beta\gamma = 18$ Given that one root is thrice the sum of the other two,let $\alpha = 3(\beta+\gamma)$. Substituting this into the sum of roots: $\alpha + \frac{\alpha}{3} = 12 \implies \frac{4\alpha}{3} = 12 \implies \alpha = 9$. Then $\beta+\gamma = 3$ and $\beta\gamma = \frac{18}{\alpha} = \frac{18}{9} = 2$. We need to find $\alpha^2+\beta^2+\gamma^2-k$. We know $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = 12^2 - 2k = 144 - 2k$. Thus,$\alpha^2+\beta^2+\gamma^2-k = 144 - 2k - k = 144 - 3k$. Since $\beta+\gamma=3$ and $\beta\gamma=2$,the roots $\beta$ and $\gamma$ are roots of $t^2-3t+2=0$,which are $1$ and $2$. Using $\alpha\beta+\beta\gamma+\gamma\alpha = k$,we get $k = \alpha(\beta+\gamma) + \beta\gamma = 9(3) + 2 = 27 + 2 = 29$. Finally,$144 - 3(29) = 144 - 87 = 57$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-12x^2+kx-18=0$ and one of them is thrice the sum of the other two roots,then $\alpha^2+\beta^2+\gamma^2-k=$

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