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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

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: go graphical | Comment 4 of 9 |
(In reply to go graphical by Larry)

Thank you for your excellent observations, Larry, which have led me to the following approach.

I f we let u=x-2, the equation becomes

sqrt(1+u) + sqrt(1-u) = u^2 + 2.

The equation is now symmetrical around u=0, and clearly has u=0 as a solution.  Thus we need only consider u>0 in seeking other possible solutions. And we need not consider u>1 as sqrt(1-u) is not real when u>1. The derivatives of the lhs and rhs are equal when u= 0, but as u increases, they have opposite sign since clearly the derivative of the rhs is positive, and that of the lhs is

(1/2)(1/sqrt(1+u) - 1/sqrt(1-u))

and 1/sqrt(1-u) > 1/sqrt(1+u). Hence the lhs and rhs diverge away from each other as u increases from 0 to 1.  u=0 is the only solution.

Edited on August 17, 2006, 12:37 am
  Posted by Richard on 2006-08-17 00:33:03

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