What is the real solution of the system x^2+y^2=16, xy=3 ?

Expert Answers

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We have the system of equations x^2 + y^2 = 16 and xy = 13 to solve for x.

xy = 3 => x = 3/y...(1)

Substitute this in x^2 + y^2 = 16

=> (3/y)^2 + y^2 = 16

=> 9 + y^4 = 16y^2

=> y^4 - 16y^2 + 9 = 0

y^2 = 16/2 + [16^2 - 36]/2

=> 8 + (sqrt 220)/2

=> 8 + sqrt 55

and y^2 = 8 - sqrt 55

y = sqrt (8 + sqrt 55) and y = sqrt(8 - sqrt 55)

x = 3/(sqrt (8 + sqrt 55)) and x = 3/(sqrt(8 - sqrt 55))

Also in (1) we could have substituted x for y. This gives us four solutions for x and y:

(sqrt (8 + sqrt 55), 3/(sqrt (8 + sqrt 55))), (sqrt (8 - sqrt 55), 3/(sqrt (8 - sqrt 55))), (3/(sqrt (8 + sqrt 55)), sqrt (8 + sqrt 55)) and (3/(sqrt (8 - sqrt 55)), sqrt (8 - sqrt 55))

The solutions of the equations are (sqrt (8 + sqrt 55), 3/(sqrt (8 + sqrt 55))), (sqrt (8 - sqrt 55), 3/(sqrt (8 - sqrt 55))), (3/(sqrt (8 + sqrt 55)), sqrt (8 + sqrt 55)) and (3/(sqrt (8 - sqrt 55)), sqrt (8 - sqrt 55))

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