Question

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

Answer 1

81

Answer 2

Let the roots be $r$, $s$, and $t$. Since they are in harmonic progression (HP), their reciprocals are in arithmetic progression (AP), so

$$\frac{2}{s} = \frac{1}{r} + \frac{1}{t} \quad \Longrightarrow \quad 2rt = s(r+t).$$

Using Vieta’s formulas for the cubic

$$kx^3 - 18x^2 - 36x + 8 = 0,$$

(dividing by $k$), we have:

$$r+s+t = \frac{18}{k}, \quad rs+rt+st = -\frac{36}{k}, \quad rst = -\frac{8}{k}.$$

Express $r+t$ and $rt$ in terms of $s$:

$$r+t = \frac{18}{k} - s,$$ $$rt = \frac{rst}{s} = -\frac{8}{k s}.$$

Plug these into the HP condition:

$$2\left(-\frac{8}{k s}\right) = s\left(\frac{18}{k}-s\right) \quad \Longrightarrow \quad -\frac{16}{k s} = \frac{18s}{k} - s^2.$$

Multiplying both sides by $k s$:

$$-16 = 18 s^2 - k s^3.$$

Rearrange:

$$k s^3 - 18s^2 - 16 = 0. \quad (1)$$

Note that $s$ is a root of the original equation. Substituting $s$ in the polynomial (after multiplying by $k$):

$$k s^3 - 18s^2 - 36s + 8 = 0. \quad (2)$$

Subtract equation (1) from (2):

$$[k s^3 - 18s^2 - 36s + 8] - [k s^3 - 18s^2 - 16] = -36s + 24 = 0.$$

Thus,

$$-36s + 24 = 0 \quad \Longrightarrow \quad s = \frac{24}{36} = \frac{2}{3}.$$

Now substitute $s = \frac{2}{3}$ into equation (1):

$$k\left(\frac{2}{3}\right)^3 - 18\left(\frac{2}{3}\right)^2 - 16 = 0.$$

Calculate:

$$k \cdot \frac{8}{27} - 18\cdot \frac{4}{9} - 16 = 0 \quad \Longrightarrow \quad \frac{8k}{27} - 8 - 16 = 0,$$ $$\frac{8k}{27} - 24 = 0 \quad \Longrightarrow \quad \frac{8k}{27} = 24,$$ $$k = 24 \times \frac{27}{8} = 81.$$