Question

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

Accepted Answer

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))

Answer

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}.

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

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