If x2−4x+1=0, then x2−1x2=

x=4±16−42=4±232=2±3x=2+3x=2=3(x−2)3=(3)3⇒x3−6x2+12x−8=3312=2−3x+1x=2+3+2−3=4x−1x=2+3−2+3=23x2−1x2=4(23)=83

If x2−4x+1=0, then x2−1x2=

x=4±16−42=4±232=2±3x=2+3x=2=3(x−2)3=(3)3⇒x3−6x2+12x−8=3312=2−3x+1x=2+3+2−3=4x−1x=2+3−2+3=23x2−1x2=4(23)=83

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If $x^{2} - 4 x + 1 = 0$ , then $x^{2} - \frac{1}{x^{2}} =$

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$8 \sqrt{3}$

$8 + \sqrt{3}$

$- 8 \sqrt{3}$

answer is A.

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Detailed Solution

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

$= \frac{4 \pm 2 \sqrt{3}}{2} = 2 \pm \sqrt{3}$

$x = 2 + \sqrt{3}$

$x = 2 = \sqrt{3}$

${(x - 2)}^{3} = {(\sqrt{3})}^{3} \Rightarrow x^{3} - 6 x^{2} + 12 x - 8 = 3 \sqrt{3}$

$\frac{1}{2} = 2 - \sqrt{3}$

$x + \frac{1}{x} = 2 + \sqrt{3} + 2 - \sqrt{3} = 4$

$x - \frac{1}{x} = 2 + \sqrt{3} - 2 + \sqrt{3} = 2 \sqrt{3}$

$x^{2} - \frac{1}{x^{2}} = 4 (2 \sqrt{3}) = 8 \sqrt{3}$

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If $x^{2} - 4 x + 1 = 0$ , then $x^{2} - \frac{1}{x^{2}} =$