The roots of the equation x^3-14x^2+56x-64=0 | vedclass

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(D) Given the cubic equation: $x^3-14x^2+56x-64=0$. By Vieta's formulas,for roots $\alpha, \beta, \gamma$: $\alpha+\beta+\gamma=14$ $\alpha\beta+\beta

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geometric progression

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(D) Given the cubic equation: $x^3-14x^2+56x-64=0$. By Vieta's formulas,for roots $\alpha, \beta, \gamma$: $\alpha+\beta+\gamma=14$ $\alpha\beta+\beta\gamma+\gamma\alpha=56$ $\alpha\beta\gamma=64$ By testing small integer values,we find that $x=2$ is a root: $2^3-14(2^2)+56(2)-64 = 8-56+112-64 = 0$. Dividing the polynomial by $(x-2)$,we get: $(x-2)(x^2-12x+32)=0$ $(x-2)(x-4)(x-8)=0$ The roots are $\alpha=2, \beta=4, \gamma=8$. Since $\frac{4}{2} = \frac{8}{4} = 2$,the roots are in Geometric Progression $(GP)$.

The roots of the equation $x^3-14x^2+56x-64=0$ are in

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