What is the real solution of the system x^2+y^2=16, xy=3 ?
- print Print
- list Cite
Expert Answers

calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
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 +...
(The entire section contains 125 words.)
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Related Questions
- Solve for x and y x^3-y^3=7 x^2+xy+y^2=7
- 2 Educator Answers
- Solve the system of equations algebraically x^2+y^2=100 x-y=2
- 1 Educator Answer
- What are x,y and z? x+y=-3 x+z=-2 xy+yz+xz=2
- 1 Educator Answer
- What are the real solutions of the equation y^2+3=13/(y^2-9) ?
- 1 Educator Answer
- How to solve the system x+y=3 and x^2/y+y^2/x=9/2?
- 1 Educator 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}.
Student Answers