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(C) Given equation: $8x^3 - 42x^2 + 63x - 27 = 0$ . . . $(i)$Since $\alpha, \beta, \gamma$ are roots of $(i)$,the product of roots is $\alpha \beta \g

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(C) Given equation: $8x^3 - 42x^2 + 63x - 27 = 0$ . . . $(i)$Since $\alpha, \beta, \gamma$ are roots of $(i)$,the product of roots is $\alpha \beta \gamma = -(\frac{-27}{8}) = \frac{27}{8}$.Given $\beta, \gamma, \alpha$ are in geometric progression,so $\gamma^2 = \beta \alpha$.Substituting this into the product of roots: $\gamma \cdot \gamma^2 = \frac{27}{8} \implies \gamma^3 = \frac{27}{8} \implies \gamma = \frac{3}{2}$.Sum of roots: $\alpha + \beta + \gamma = \frac{42}{8} = \frac{21}{4}$.Since $\alpha + \beta = \frac{21}{4} - \frac{3}{2} = \frac{15}{4}$ and $\alpha \beta = \gamma^2 = \frac{9}{4}$,$\alpha$ and $\beta$ are roots of $t^2 - \frac{15}{4}t + \frac{9}{4} = 0 \implies 4t^2 - 15t + 9 = 0 \implies (4t - 3)(t - 3) = 0$.Thus,$t = \frac{3}{4}$ or $t = 3$. Given $\beta The expression is $\frac{3}{2}x^2 + 4(\frac{3}{4})x + 3 = \frac{3}{2}x^2 + 3x + 3$.The extreme value of $ax^2 + bx + c$ is $-\frac{D}{4a} = \frac{4ac - b^2}{4a}$.Here $a = \frac{3}{2}, b = 3, c = 3$,so the extreme value is $\frac{4(\frac{3}{2})(3) - (3)^2}{4(\frac{3}{2})} = \frac{18 - 9}{6} = \frac{9}{6} = \frac{3}{2}$.

$\alpha, \beta, \gamma$ are the roots of the equation $8x^3 - 42x^2 + 63x - 27 = 0$. If $\beta < \gamma < \alpha$ and $\beta, \gamma, \alpha$ are in geometric progression,then the extreme value of the expression $\gamma x^2 + 4\beta x + \alpha$ is

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