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If $x^{2} - 4 x - 1 = 0$ and $x > 0$ , then $x^{3} - 6 x^{2} + 12 x =$

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- 8 - 5 \sqrt{5}

8 + 5 \sqrt{5}

8 - 5 \sqrt{5}

- 8

answer is C.

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Detailed Solution

x^{2} - 4 x - 1 = 0

and

x > 0

then

x^{3} - 6 x^{2} + 12 x

$x = \frac{4 \pm \sqrt{16 + 4}}{2}$ $= \frac{4 \pm 2 \sqrt{5}}{2} = 2 + \sqrt{5}$

$x - 2 = \sqrt{5}$

${(x - 2)}^{3} = {(\sqrt{5})}^{3}$

$x^{3} - 6 x^{2} + 12 x - 8 = 5 \sqrt{5}$

$x^{3} - 6 x^{2} + 12 x - 8 + 5 \sqrt{5} = 0$

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