x and y error

Algebra Level 3

Find the value of


The answer is 584.

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1 solution

It is given that x = 3 + 2 2 \space x = \sqrt{3}+2\sqrt{2}\space and y = 3 2 2 \space y = \sqrt{3}-2\sqrt{2}

x + y = 2 3 \Rightarrow x + y = 2\sqrt{3} \space and x y = ( 3 ) 2 ( 2 2 ) 2 = 5 \space xy = (\sqrt{3})^2 - (2\sqrt{2})^2 = -5

( x + y ) 4 = x 4 + 4 x 3 y + 6 x 2 y 2 + 4 x y 3 + y 4 (x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4

x 4 + y 4 + 6 x 2 y 2 = ( x + y ) 4 4 x 3 y 4 x y 3 = ( x + y ) 4 4 x y ( x 2 + y 2 ) \Rightarrow x^4 + y^4 + 6x^2y^2 = (x+y)^4 - 4x^3y - 4xy^3 = (x+y)^4 - 4xy (x^2 + y^2)

= ( x + y ) 4 4 x y [ ( x + y ) 2 2 x y ] = ( 2 3 ) 4 4 ( 5 ) [ ( 2 3 ) 2 2 ( 5 ) ] \quad = (x+y)^4 - 4xy [(x+y)^2-2xy] = (2\sqrt{3})^4 - 4(-5)[(2\sqrt{3})^2 -2(-5)]

= 144 + 20 ( 22 ) = 584 \quad = 144 + 20(22) = \boxed{584}

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