If one root of the equation x^3-9x^2+26x-24=0 | vedclass

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(A) Let the roots of the equation $x^3-9x^2+26x-24=0$ be $\alpha, \beta, \gamma$ such that $\alpha = 2\beta$.From the relations between roots and coef

Vedclass · Accepted Answer

$72$

Vedclass · Answer

(A) Let the roots of the equation $x^3-9x^2+26x-24=0$ be $\alpha, \beta, \gamma$ such that $\alpha = 2\beta$.From the relations between roots and coefficients:$1$) $\alpha + \beta + \gamma = 9$ $\Rightarrow 3\beta + \gamma = 9$ $\Rightarrow \gamma = 9 - 3\beta$$2$) $\alpha\beta\gamma = 24$ $\Rightarrow (2\beta)\beta\gamma = 24$ $\Rightarrow \beta^2\gamma = 12$Substituting $\gamma$ in the second equation:$\beta^2(9 - 3\beta) = 12$ $\Rightarrow 9\beta^2 - 3\beta^3 = 12$ $\Rightarrow \beta^3 - 3\beta^2 + 4 = 0$Factoring the cubic equation: $(\beta + 1)(\beta - 2)^2 = 0$.Thus,$\beta = -1$ or $\beta = 2$.Case $1$: If $\beta = -1$,then $\alpha = 2\beta = -2$. The sum of cubes is $\alpha^3 + \beta^3 = (-2)^3 + (-1)^3 = -8 - 1 = -9$.Case $2$: If $\beta = 2$,then $\alpha = 2\beta = 4$. The sum of cubes is $\alpha^3 + \beta^3 = 4^3 + 2^3 = 64 + 8 = 72$.Since $72$ is an option,the correct answer is $72$.

If one root of the equation $x^3-9x^2+26x-24=0$ is twice the other,then the sum of the cubes of those two roots is

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