Given x+y=3*square root3 and x*y=3 calculate x^2+y^2 and x^4+y^4.

Expert Answers

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We are given that x + y = 3*sqrt 3 and x*y = 3

x^2 + y^2 = (x + y)^2 - 2xy

=> (3*sqrt 3)^2 - 2*3

=> 9*3 - 6

=> 27 - 6

=> 21

x^4 + y^4 = (x^2 + y^2)^2 - 2x^2*y^2

=> 21^2 - 2*(x*y)^2

=> 441 - 2*3^2

=> 441 - 2*9

=> 441 - 18

=> 423

The required values of x^2 + y^2  and x^4 + y^4 are 21 and 423 respectively.

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