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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)

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Solution a solution | Comment 5 of 9 |

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

sqrt(1+(x-2))+sqrt(1-(x-2))=(x-2)^2+2

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

sqrt(1+y)+sqrt(1-y)=y^2+2

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
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