Solve the following equation that has equal roots:
$x^{6} - 2 x^{5} - 4 x^{4} + 12 x^{3} - 3 x^{2} - 18 x + 18 = 0$ .

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Solution

Give equation, $x^{6} - 2 x^{5} - 4 x^{4} + 12 x^{3} - 3 x^{2} - 18 x + 18 = 0$

Consider $f (x) = x^{6} - 2 x^{5} - 4 x^{4} + 12 x^{3} - 3 x^{2} - 18 x + 18$

$∴ f^{'} (x) = 6 x^{5} - 10 x^{4} + 16 x^{3} + 36 x^{2} - 6 x + 18$

Now, HCF of $f (x)$ and $f^{'} (x)$ is $(x^{2} - 3)$ . Hence $x = \pm \sqrt{3}$ is a double root of $f (x) = 0$

$f (x)$ can be factored as $(x^{2} - 3)^{2} (x^{2} - 2 x + 2)$

$∴$ roots of the given equation are $\pm \sqrt{3}, \pm \sqrt{3}, 1 \pm i \sqrt{1}$

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