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 A Real Number problem (Posted on 2006-08-16)
Given that x is a real number, determine all possible solutions of:
√(x-1) + √(3-x) = x² - 4x + 6

 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

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