Solving to Find x - Not Getting ALL Solutions

[eutas1]

Please refer to the attachments for the question (focus part inside red outline) and my working out:

How come I am only getting one of the solutions (x = 1)?? I know that x = 0 and x= - 1 turn out to be invalid solutions (using previous calculations not shown in this post) and you end up using x = 1 anyway but I'm not sure what I am doing wrong in that I am not getting ALL solutions?

 

[Deleted member 4993]

Please refer to the attachments for the question (focus part inside red outline) and my working out:

How come I am only getting one of the solutions (x = 1)?? I know that x = 0 and x= - 1 turn out to be invalid solutions (using previous calculations not shown in this post) and you end up using x = 1 anyway but I'm not sure what I am doing wrong in that I am not getting ALL solutions?

In the second line of your solution, you had multiplied and divided by 'x'. That discards the solution x = 0.

in fourth line you multiplied by (x+1) → that discarded the solution x = -1

 

[Dr.Peterson]

I have two comments.

First, I wouldn't solve it either your way or theirs:



They cross-multiplied, which is wasteful; I'd just multiply by 2x^2 and get 2x(x+1)^2 = (x+1)^3; so I would never see the extraneous solution x = 0.

Second, once I got here, I would not divide, but subtract and factor (which is the way to avoid your mistake):

2x(x+1)^2 - (x+1)^3 = 0​

(x+1)^2(2x - (x+1)) = 0​

(x+1)^2(x-1) = 0​

This gives x = -1 or 1. Both are valid solutions to this equation; -1 is invalid due to some earlier step in the work.