If the roots of the equation 16x^3 - 44x^2 + | vedclass

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(B) The given equation is $16x^3 - 44x^2 + 36x - 9 = 0$. Let the roots be $\alpha, \beta, \gamma$ in harmonic progression ($H$.$P$.). Then $\frac{1}{\

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$\frac{3}{2}$

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(B) The given equation is $16x^3 - 44x^2 + 36x - 9 = 0$. Let the roots be $\alpha, \beta, \gamma$ in harmonic progression ($H$.$P$.). Then $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$ are in arithmetic progression ($A$.$P$.). This implies $\frac{2}{\beta} = \frac{1}{\alpha} + \frac{1}{\gamma}$. From the properties of roots,$\sum \alpha\beta = \frac{36}{16} = \frac{9}{4}$ and $\alpha\beta\gamma = \frac{9}{16}$. Also,$\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} = \frac{9/4}{9/16} = 4$. Substituting $\frac{1}{\alpha} + \frac{1}{\gamma} = \frac{2}{\beta}$,we get $\frac{2}{\beta} + \frac{1}{\beta} = 4$ $\Rightarrow \frac{3}{\beta} = 4$ $\Rightarrow \beta = \frac{3}{4}$. Now,$\alpha + \gamma = \frac{44}{16} - \frac{3}{4} = \frac{11}{4} - \frac{3}{4} = 2$ and $\alpha\gamma = \frac{9/16}{3/4} = \frac{3}{4}$. Solving $t^2 - 2t + \frac{3}{4} = 0$,we get $4t^2 - 8t + 3 = 0 \Rightarrow (2t - 1)(2t - 3) = 0$. Thus,the roots are $\frac{1}{2}, \frac{3}{4}, \frac{3}{2}$. The greatest root is $\frac{3}{2}$.

If the roots of the equation $16x^3 - 44x^2 + 36x - 9 = 0$ are in harmonic progression,then its greatest root is

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