Question

If x=2+root3, find the value of x^2 + 1/x^2

Answer 1

$x=2+\sqrt3\\\\\frac1{x^2}=\frac1{(2+\sqrt3)^2}=\frac{(2-\sqrt3)^2}{(2+\sqrt3)^2(2-\sqrt3)^2}=\\\\\frac{(2-\sqrt3)^2}{[(2+\sqrt3)(2-\sqrt3)]^2}=\frac{(2-\sqrt3)^2}{[2^2-(\sqrt3)^2]^2}=\\\\\frac{(2-\sqrt3)^2}{(4-3)^2}=\frac{(2-\sqrt3)^2}{1}=(2-\sqrt3)^2\\\\x^2+\frac1{x^2}=(2+\sqrt3)^2+(2-\sqrt3)^2=\\\\4+2\sqrt3+(\sqrt3)^2+4-2\sqrt3+(\sqrt3)^2=4+3+4+3=14$

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


Thanks for the question!

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