Question 66123: I need to solve for all real roots
(sqrt)z-(sqrt)(z-1)=1
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! 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
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