Further Geometric Numbers (Posted on 2006-11-24)

	Submitted by K Sengupta
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Let x1,x2,x3 and x4 correspond to the four roots of the given equation and let us suppose without any loss of generality that these roots describe a geometric sequence in the order (x1, x3, x2, x4). We write Vieta's relations in terms of s = x1+x2; t = x3+x4; p = x1*x2 and q = x3*x4. This gives: s+t = 15 p+q+st = 70 pt+qs = 120 pq = m Since (x1,x3, x4, x2) describe a geometric sequence, it follows that x1*x2 = x3*x4, and hence p = q. Accordingly, 120 = pt+ qs = p(s+t)= 15p, and therefore, p=8. Then, m = 64 and st = 70 - 16 = 54. The numbers s and t are roots of the equation x^2 - 15x+54 = 0. Hence, s =6 and t =9 or vice-versa. Consequently, it follows that, the roots of the polynomial are 1,2,4,8. For an alternate methodology, refer to the solution posted by Bractals in this location.

	Subject	Author	Date
	Solution	Bractals	2006-11-24 06:56:50