Solve $6 x^{3}$ - $11 x^{2}$ $+ 6 x - 1 = 0$ , if its roots are in harmonic progression.

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Solution

Consider the given equation,

$6 x^{3} - 11 x^{2} + 6 x - 1 = 0$

Put, $x = 1$ we get

${6.1}^{3} - {11.1}^{2} + 6.1 - 1 = 0$

$0 = 0$

Hence, $x = 1$ $\Rightarrow x - 1 = 0$ is zeroes os given equation.

Now

$x - 1 6 x^{2} - 5 x + 1 ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯) 6 x^{3} - 11 x^{2} + 6 x - 1$

$- (6 x^{3} - 6 x^{2}) - -------------- -$

$- 5 x^{2} + 6 x - 1$

$- (- 5 x^{2} + 5 x) - --------------- -$

$x - 1$

$- (x - 1) - -------- -$

$0$

Now,

$6 x^{2} - 5 x + 1 = 0$

$6 x^{2} - 3 x - 2 x + 1 = 0$

$3 x (2 x - 1) - (2 x - 1) = 0$

$(2 x - 1) (3 x - 1) = 0$

So,

$6 x^{3} - 11 x^{2} + 6 x - 1 = (2 x - 1) (3 x - 1) (x - 1) = 0$

Hence, $x = 1, \frac{1}{2}, \frac{1}{3}$ in H.P.

Hence, this is the answer.

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