First of all x belongs to [1,3]. The equation can be written as
sqrt(1+(x2))+sqrt(1(x2))=(x2)^2+2
Let x2=y, y belongs to [1,1]. So, the equation becomes
sqrt(1+y)+sqrt(1y)=y^2+2
Let sqrt(1+y)=a and sqrt(1y)=b. Then a*b=sqrt(1y^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^22P=2 and S+P^2=3. Let S=3P^2. Then we obtain the equation in P: P^46P^22P+7=0 (1). A solution is P=1. Other solution can be for the equation P^3+P^25P7=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 20060817 03:50:54 