Question 66123
I need to solve for all real roots

(sqrt)z-(sqrt)(z-1)=1

First, we'll square both sides to start getting rid of the radicals:

[sqrt(z)-sqrt(z-1)]^2=(sqrt(z))^2-2(sqrt(z))(sqrt(z-1))+(sqrt(z-1))^2

so we have:

z-2sqrt((z)(z-1))+z-1=1 which equals

2z-1-2sqrt((z)(z-1))=1  Now, we'll subtract 2z and -1 from both sides:

-2sqrt((z)(z-1))=1-2z+1=

-2sqrt((z)(z-1))=2-2z  divide both sides by -2


sqrt((z)(z-1))=z-1  square both sides again:

(z)(z-1)=(z-1)(z-1)  


Note:  We are to this point and the tendency is to divide both sides by (z-1) and that would be a BIG mistake.  Why??????????????

(z)(z-1)=(z-1)(z-1)=
z^2-z=z^2-2z+1 subtract z^2 and -2z from both sides:

z^2-z^2-z+2z=1


z=1 is the only real solution:


AN IMPORTANT LESSON HERE:

z-1=0  AND YOU CAN'T DIVIDE BY 0



Hope this helps-----ptaylor