If alpha, beta, gamma are the roots of the eq | vedclass

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(C) Let the roots be $\alpha = \frac{a}{r}, \beta = a, \gamma = ar$. Since they are in geometric progression,the product of the roots is given by $\fr

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) Let the roots be $\alpha = \frac{a}{r}, \beta = a, \gamma = ar$. Since they are in geometric progression,the product of the roots is given by $\frac{a^3}{r} \cdot r = a^3 = \frac{24}{3} = 8$. Thus,$a = 2$. The sum of the roots is $\frac{a}{r} + a + ar = \frac{26}{3}$. Substituting $a = 2$,we get $\frac{2}{r} + 2 + 2r = \frac{26}{3}$. Dividing by $2$,we get $\frac{1}{r} + 1 + r = \frac{13}{3}$,which simplifies to $\frac{1}{r} + r = \frac{10}{3}$. Multiplying by $3r$,we get $3r^2 - 10r + 3 = 0$. Factoring gives $(3r - 1)(r - 3) = 0$,so $r = 3$ or $r = \frac{1}{3}$. Since $\alpha  1$,so $r = 3$. The roots are $\alpha = \frac{2}{3}, \beta = 2, \gamma = 6$. Finally,$3\alpha + 2\beta + \gamma = 3(\frac{2}{3}) + 2(2) + 6 = 2 + 4 + 6 = 12$.

If $\alpha, \beta, \gamma$ are the roots of the equation $3x^3-26x^2+52x-24=0$ such that $\alpha, \beta, \gamma$ are in geometric progression and $\alpha < \beta < \gamma$,then $3\alpha + 2\beta + \gamma =$

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