(In reply to
re: Solution by armando)
I agree.
I have: P^2 + R^2 + P*R(Q^2 + R^2  Q*R) = (P+Q) (PQ+R) = 14425 = 119 = 13(PQ+R), so (PQ+R) = 119/13.
Then, since P + Q=13, R= 288/132P [1] and we can substitute:
P^2 + (288/132P)^2 + P*(288/132P) = 144, so 169P^23744P+27648 = 8112, or 169P^23744P+19536 = 0;
P = 144/1320*3^(1/2)/13,or P= 144/13+20*3^(1/2)/13, but the second of these exceeds 13, implying that Q is negative, contrary to the stipulation of the problem.
But since twice 144/13 is just 288/13, from [1], R is simply 40*3^(1/2)/13
It's just solving the quadratic that needs care.
Edited on June 12, 2016, 3:52 am

Posted by broll
on 20160612 03:52:07 