Question

if x=2-root3,find the value of x2+1/xandx2-1/x2

Answer 1

If x=2-√3

By rationalization,

1/x=2+√3

Then,

x²+1/x

=(2-√3)²+(2+√3)

=4+3-4√3+4+√3

=11-3√3=(x²+1/x)

And

x²-1/x²

=(2-√3)²+(2+√3)²

=4+3-4√3+4+3+4√3

=7+7

=14

Answer 2

Hey there !


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Given :


x = 2 - √3


1/x = 1/ 2 - √3 × 2 + √3 / 2 + √3


1/x = 2 + √3 / 2² - √3²


1/x = 2 + √3 / 4 - 3


1/x = 2 + √3


Now,


x + 1/x = 2 - √3 + 2 + √3


x + 1/x = 2 + 2


x + 1/x = 4


Squaring both sides ;


( x + 1/x)² = (4)²


=> x² + 1/x² + 2 = 16


=> x² + 1/x² = 16 - 2


=> x² + 1/x² = 14

Now,

=> x² = ( 2 - √3 ) ²

=> x² = 2² + √3² - 2 * 2 * √3

=> x² = 4 + 3 - 4√3

=> x² = 7 - 4√3

And,

=> ( 1 / x² ) = ( 2 + √3 ) ²

=> ( 1 / x² ) = 2² + √3² + 2 * 2 * √3

=> ( 1 / x² ) = 4 + 3 + 4√3

=> ( 1 / x² ) = 7 + 4√3


So,

x² - 1 / x² = 7 - 4√3 - 7 - 4√3

=> x² - 1 / x² = 0
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Thanks for the question!

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