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

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### 2 Answers

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))**

We'll recognize the type of system: symmetrical system.

The property of this kind of system is if interchanging the x and y values, the equations of the system are verified.

We'll write x^2 + y^2 = (x+y)^2 - 2xy

We'll note x+y=s and xy=p

We'll re-write the equations of the system in s and p:

s^2 - 2p = 16

p = 3

We'll substitute the value of p in the 1st equation:

s^2 - 6 = 16

s^2 = 16 + 6

s^2 = 22

s1 = sqrt 22 and s2 = -sqrt 22

We'll find out x and y:

x^2 - sqrt22*x + 3 = 0

x1 = [sqrt22 + sqrt(22 - 12)]/2

x1 = (sqrt22 + sqrt10)/2

x2 = (sqrt22 - sqrt10)/2

x^2 + sqrt22*x + 3 = 0

x3 = (-sqrt22 + sqrt10)/2

x4 = (-sqrt22 - sqrt10)/2

**The solutions of the symmetrical system are: {(sqrt22 + sqrt10)/2 ; (sqrt22 - sqrt10)/2} ; {(sqrt22 - sqrt10)/2 ; (sqrt22 + sqrt10)/2} ; {(-sqrt22 + sqrt10)/2 ; (-sqrt22 - sqrt10)/2} ; {****(-sqrt22 - sqrt10)/2 ; (-sqrt22 + sqrt10)/2****}.**