The real root of the equation $x^{3} - 6 x + 9 = 0$ is

- 6

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- 9

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6

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- 3

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Solution

The correct option is B $- 3$
$x^{3} - 6 x + 9 = 0$ ,

Let $x = - 3$ ; therefore
$- 27 + 18 + 9 = 0$ .

Hence $x + 3$ is a factor of $x^{3} - 6 x + 9 = 0$ .

Thus the equation can be written as

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

$x^{2} - 3 x + 3 = 0$ and $x + 3 = 0$ .

Consider $x^{2} - 3 x + 3 = 0$ ,

$b^{2} - 4 a c$

$= 9 - 12$

$= - 3$

Hence $b^{2} - 4 a c < 0$ . Hence imaginary roots.

Therefore, the only real root is $x = - 3$ .

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