Each of P, Q and R is a positive real number satisfying this system of equations:

P + Q = 13, and:
Q² + R² - Q*R = 25, and:
P² + R² + P*R = 144

Find R.

(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) (P-Q+R) = 144-25 = 119  = 13(P-Q+R), so (P-Q+R) = 119/13.

Then, since P + Q=13, R= 288/13-2P  [1] and we can substitute:

P^2 + (288/13-2P)^2 + P*(288/13-2P) = 144, so 169P^2-3744P+27648 = 8112, or 169P^2-3744P+19536 = 0;

P = 144/13-20*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 2016-06-12 03:52:07