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

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### 2 Answers

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.**

We'll complete the sum of squares, x^2 + y^2, by adding 2xy and creating a perfect square.

x^2 + 2xy + y^2

Since we've added the amount 2xy, we'll subtract it. In this way, the result of the sum of squares, will remain unchanged.

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

We'll substitute x + y and xy by their values given in enunciation:

x^2 + y^2 = (3sqrt3)^2 - 2*3

x^2 + y^2 = 27 - 6

x^2 + y^2 = 21

Now, we'll create the perfect squares for:

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

x^4 + y^4 = 21^2 - 2*3^2

x^4 + y^4 = 441 - 18

x^4 + y^4 = 423

**So, the requested sums are: x^2 + y^2 = 21 and x^4 + y^4 = 423.**