A Real Number problem (Posted on 2006-08-16)

	Submitted by K Sengupta
Rating: 4.0000 (2 votes)
Solution:	(Hide)
Clearly, 1<= x< =3, since x is a real number. Otherwise, one of the two expressions in the LHS would be imaginary. Let, y = sqrt(x-1) + sqrt(3-x) or, y2 = 2 + 2*sqrt {(x-1)(3-x)} = 2 + 2*sqrt {1 - (x-2)^2} < = 2+2*1 = 4 or, |y| < =2-------(#) Now, x2- 4x + 6 = (x-2)2+2 >=2--------(##) Now, (#)and (##) is feasible only when : x2- 4x + 6 = 2 or, x = 2. Consequently, x =2 is the only feasible solution to the given problem.

	Subject	Author	Date
	Solution	Danish Ahmed Khan	2012-10-26 05:04:12
	It's easier than you make it	Peter	2006-12-11 20:56:52
	re(3): go graphical	Richard	2006-08-17 22:58:24
	re(2): go graphical	Gamer	2006-08-17 20:30:55
	a solution	Stefan	2006-08-17 03:50:54
	re: go graphical	Richard	2006-08-17 00:33:03
	go graphical	Larry	2006-08-16 21:39:45
	my solution	Daniel	2006-08-16 16:21:36
	An idea	Old Original Oskar!	2006-08-16 13:36:40