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(D) Given the cubic equation $x^3-9 x^2+k=0$,where the roots are $\alpha, \beta, 2 \beta$. Using Vieta's formulas: Sum of roots: $\alpha + \beta + 2 \

Vedclass · Accepted Answer

$54$

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(D) Given the cubic equation $x^3-9 x^2+k=0$,where the roots are $\alpha, \beta, 2 \beta$. Using Vieta's formulas: Sum of roots: $\alpha + \beta + 2 \beta = 9 \implies \alpha + 3 \beta = 9$  $(i)$ Sum of roots taken two at a time: $\alpha \beta + \beta(2 \beta) + 2 \beta(\alpha) = 0$ (since the coefficient of $x$ is $0$) $\alpha \beta + 2 \beta^2 + 2 \alpha \beta = 0 \implies 3 \alpha \beta + 2 \beta^2 = 0$ $\beta(3 \alpha + 2 \beta) = 0$ Since $k 
eq 0$,the product of roots $\alpha \cdot \beta \cdot 2 \beta = -k 
eq 0$,so $\beta 
eq 0$. Thus,$3 \alpha + 2 \beta = 0 \implies \alpha = -\frac{2 \beta}{3}$  $(ii)$ Substitute $(ii)$ into $(i)$: $-\frac{2 \beta}{3} + 3 \beta = 9$ $\frac{-2 \beta + 9 \beta}{3} = 9$ $\frac{7 \beta}{3} = 9 \implies 7 \beta = 27$ Therefore,$14 \beta = 2 	imes (7 \beta) = 2 	imes 27 = 54$. Thus,the correct option is $(d)$.

If $\alpha, \beta, 2 \beta$ are the real roots of the equation $x^3-9 x^2+k=0$ and $k \in R-\{0\}$,then $14 \beta=$

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