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A Real Number problem (Posted on 2006-08-16) Difficulty: 2 of 5
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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Some ThoughtsSolutionDanish Ahmed Khan2012-10-26 05:04:12
It's easier than you make itPeter2006-12-11 20:56:52
re(3): go graphicalRichard2006-08-17 22:58:24
re(2): go graphicalGamer2006-08-17 20:30:55
Solutiona solutionStefan2006-08-17 03:50:54
re: go graphicalRichard2006-08-17 00:33:03
Solutiongo graphicalLarry2006-08-16 21:39:45
my solutionDaniel2006-08-16 16:21:36
Some ThoughtsAn ideaOld Original Oskar!2006-08-16 13:36:40
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