All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
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.)
Solution a solution | Comment 5 of 9 |

First of all x belongs to [1,3]. The equation can be written as


Let x-2=y, y belongs to [-1,1]. So, the equation becomes


Let sqrt(1+y)=a and sqrt(1-y)=b. Then a*b=sqrt(1-y^2) and y^2=1-(a^2)*(b^2). a and b verify the next two equations:

a^2+b^2=2 and a+b+(a^2)*(b^2)=3.

Let a+b=S and a*b=P. The equations in a and b can be written as S^2-2P=2 and S+P^2=3. Let S=3-P^2. Then we obtain the equation in P: P^4-6P^2-2P+7=0 (1). A solution is P=1. Other solution can be for the equation P^3+P^2-5P-7=0. Using the Rolle's criteria for the number of real solutions of the equation, we find that the equation has another solution that belongs to (2,3). But, if 2<P<3 then 4<P^2<9 and results that S<0, so a<0 and b<0, which is impossible. So the unique convenient solution of the equation (1) is P=1. Then S=2 and a=b=1. Results that y=0, so x=2.

  Posted by Stefan on 2006-08-17 03:50:54
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2021 by Animus Pactum Consulting. All rights reserved. Privacy Information