Determine all possible triple(s) (p, q, r) of rational numbers that satisfy the following system of simultaneous equations:

p²(q+r) = 5, q²(r + p) = 8, and r²(p + q) = 9

p²(q+r)=5
q²(r+p)=8
r²(p+q)=9
Multiply these 3 equations
(pqr)²(p+q)(q+r)(r+p)=360
(pqr)²(pq+pr+q²+qr)(r+p)=360
(pqr)²(pqr+p²q+p²r+pr²+q²r+q²p+qr²+pqr)=360
Substitute those prev values in the above equation
(pqr)²(2pqr+5+8+9)=360
(pqr)²(2pqr+22)=360
=>(pqr)²(pqr+11)=180
Let pqr=x => x³+11x²-180=0
The only rational solution to this equation is x=-6
=>pqr=-6 --- (a)
Substitute this in 1st 3 equations to get
1/r+1/q=-5/6p  --- (1)
1/p+1/r=-4/3q  --- (2)
1/p+1/q=-3/2r  --- (3)

(1)-(2) => 1/q-1/p=-5/6p+4/3q
=> -1/6p=1/3q => q=-2p -- (4)

(1)-(3) => 1/r-1/p=-5/6p+3/2r
=> -1/6p = 1/2r => r=-3p -- (5)

Sub (4) and (5) in eq(a)
6p³=-6 => p=-1, q=2,r=3
The only solution to given set of equations is (-1,2,3)

Edited on May 20, 2008, 2:37 am

Posted by Praneeth on 2008-05-19 12:55:29