If the roots of the equation k x^3 - 18 x^2 - | vedclass

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(C) Given the equation $k x^3 - 18 x^2 - 36 x + 8 = 0$. Let the roots be $\alpha, \beta, \gamma$ which are in harmonic progression $(HP)$. This implie

Vedclass · Accepted Answer

$81$

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(C) Given the equation $k x^3 - 18 x^2 - 36 x + 8 = 0$. Let the roots be $\alpha, \beta, \gamma$ which are in harmonic progression $(HP)$. This implies that $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$ are in arithmetic progression $(AP)$. Replacing $x$ with $\frac{1}{x}$ in the original equation,we get $k(\frac{1}{x})^3 - 18(\frac{1}{x})^2 - 36(\frac{1}{x}) + 8 = 0$,which simplifies to $8 x^3 - 36 x^2 - 18 x + k = 0$. Let the roots of this new equation be $a-d, a, a+d$. From the sum of roots,$(a-d) + a + (a+d) = -(\frac{-36}{8}) = \frac{9}{2}$ $\Rightarrow 3a = \frac{9}{2}$ $\Rightarrow a = \frac{3}{2}$. Since $a = \frac{3}{2}$ is a root of $8 x^3 - 36 x^2 - 18 x + k = 0$,we substitute it: $8(\frac{3}{2})^3 - 36(\frac{3}{2})^2 - 18(\frac{3}{2}) + k = 0$. $8(\frac{27}{8}) - 36(\frac{9}{4}) - 27 + k = 0$. $27 - 81 - 27 + k = 0$. $k - 81 = 0 \Rightarrow k = 81$.

If the roots of the equation $k x^3 - 18 x^2 - 36 x + 8 = 0$ are in harmonic progression,then $k =$

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