Take Second Degree, Solve For Real (Posted on 2007-04-27)

	Submitted by K Sengupta
Rating: 3.0000 (1 votes)
Solution:	(Hide)
Multiplying the first equation by m and the second equation by n and subtracting the second result from the first we obtain: 15m^3 - 3*m^2*n + 15m*n^2 - 3*n^3 = 0 Or, 5m^3 - m^2*n + 5m*n^2- n^3 = 0 Or, (5m-n)(m^2 + n^2) = 0; Now, if, m^2 + n^2 = 0 then m is imaginary whenever n is real and vice versa.This is a contradiction. Accordingly, 5m = n, and substituting this in the second equation, we obtain: 5*m^3 = 95*m^2; so that : Either, m = 0; giving n = 0 Or, m= 19; giving n = 95 Consequently, (m, n) = (0,0); (19, 95) are the only possible solutions to the given problem.

	Subject	Author	Date
	re: Solution	Bractals	2007-04-27 20:10:32
	Solution	hoodat	2007-04-27 18:08:36
	Solution	Bractals	2007-04-27 12:05:06
	solution	xdog	2007-04-27 10:59:26