If x=2−3, then x4−4x3+2x2−4x−5=

x=2−3 1x=12−3=2+3 ∴x+1x=4⋯⋯(1) Squaring on both sides x2+1x2=14⋯⋯(2) To find x4−4x3+2x2−4x−5 x4−4x3+2x2−4x+1−6 x2(x2+1x2)−4x2(x+1x)+2x2−6 x2(14)−4x2(4)+2x2−6 14x2−16x2+2x2−6=-6

If x=2−3, then x4−4x3+2x2−4x−5=

x=2−3 1x=12−3=2+3 ∴x+1x=4⋯⋯(1) Squaring on both sides x2+1x2=14⋯⋯(2) To find x4−4x3+2x2−4x−5 x4−4x3+2x2−4x+1−6 x2(x2+1x2)−4x2(x+1x)+2x2−6 x2(14)−4x2(4)+2x2−6 14x2−16x2+2x2−6=-6

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If $x = 2 - \sqrt{3}$ , then $x^{4} - 4 x^{3} + 2 x^{2} - 4 x - 5 =$

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$- 8$

$- 4$

$- 6$

answer is D.

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Detailed Solution

$x = 2 - \sqrt{3} \frac{1}{x} = \frac{1}{2 - \sqrt{3}} = 2 + \sqrt{3} ∴ x + \frac{1}{x} = 4 \dots \dots (1) S q u a r i n g o n b o t h s i d e s x^{2} + \frac{1}{x^{2}} = 14 \dots \dots (2) T o f i n d x^{4} - 4 x^{3} + 2 x^{2} - 4 x - 5 x^{4} - 4 x^{3} + 2 x^{2} - 4 x + 1 - 6 x^{2} (x^{2} + \frac{1}{x^{2}}) - 4 x^{2} (x + \frac{1}{x}) + 2 x^{2} - 6 x^{2} (14) - 4 x^{2} (4) + 2 x^{2} - 6 14 x^{2} - 16 x^{2} + 2 x^{2} - 6 = - 6$