Symmetric Functions II
This problem is from Alexander Karabegov.
Solve the following system of equations (you may want to look back at the previous Problem of the Week):
- x3 + y3 = 9
- 5xy = 2x2 + 2y2.
Please submit all solutions to mathpotw@acu.edu by 5:00 PM on Thursday.
Correct solutions were submitted by: Ricky Kagoro Lumala
There are several approaches that will work. Following are two of them.
Approach I . We use the results from the previous problem of the week. So we assume that there is a polynomial z2 + p z + q with two real roots x and y. Then we have found that
x3 + y3 = –p3 + 3pq,
x2 + y2 = p2 – 2q, and
xy = q.
Substituting these identities into the given system of equations yields (after some simplification) an equivalent system of equations:
–p3 + 3pq = 9 and
9q = 2p2.
This system is easy to solve: p = -3 and q = 2.
Therefore the corresponding quadratic equation is z2 – 3z + 2 = 0. The roots of this equation are x = 1 or x = 2. Therefore the solutions of the given equations are:
(x, y) = (1, 2) and (x, y) = (2,1).
Approach II . Divide both sides of the equation 5xy = 2x2 + 2y2 by xy to get
5 = 2(x/y) + 2(y/x). Now let a = x/y which, after rearranging, gives the quadratic equation 2a2 – 5a + 2 = (2a – 1)(a – 2) = 0. So a = 1/2 or a = 2.
This in turn gives 2x = y or x = 2y, respectively. Substituting these into the equation
x3 + y3 = 9 easily give the same solutions as in the previous approach.
