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

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Aicharya · Accepted Answer

The given equation is $16 x^3-44 x^2+36 x-9=0$ Let $\alpha, \beta$ and $\gamma$ are in H.P. $\Rightarrow \frac{3}{\beta}=\frac{1}{\alpha}+\frac{1}{\gamma}+\frac{1}{\beta} \Rightarrow \frac{3}{\beta}=\frac{\alpha \beta+\beta \gamma+\gamma \alpha}{\alpha \gamma \beta}$ $\Rightarrow \frac{3}{\beta}=\frac{36 / 16}{9 / 16}=4 \Rightarrow \beta=\frac{3}{4}$ Now, $\alpha+\beta+\gamma=\frac{11}{4} \Rightarrow \alpha+\gamma=\frac{11}{4}-\frac{3}{4}=2$ ...(i) And $\alpha \beta \gamma=\frac{9}{16} \Rightarrow \alpha \gamma=\frac{9}{16} 	imes \frac{4}{3}=\frac{3}{4}$...(ii) Solving eqn. (i) and (ii) : $\alpha=\frac{3}{2}, \gamma=\frac{1}{2}$ $	herefore$ Greatest root is $\alpha=\frac{3}{2}$.

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

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