Elementary Equation solving mind-block

[truewt]

Hi I have some problems with really basic equations (really elementary)

Let's say in order to solve an equation $$f(x)=0$$, we multiply the equation by $$x$$. Therefore we conclude that x can never be =0. But what if at the end step we conclude that $$x=0$$ (maybe along with other solutions)? Do we reject the answer and accept the others? Or is our method of solving the equation incorrect?

I do know that it is quite impossible for you to arrive at $$x=0$$ after you multiply $$x$$ throughout in order to solve the equation, as that would mean you had introduced an unnecessary common multiple into the equation..

 

[HallsofIvy]

Hi I have some problems with really basic equations (really elementary)

Let's say in order to solve an equation f(x)=0, we multiply the equation by x. Therefore we conclude that x can never be =0.

Why would we conclude that?

But what if at the end step we conclude that x=0 (maybe along with other solutions)? Do we reject the answer and accept the others? Or is our method of solving the equation incorrect?

I do know that it is quite impossible for you to arrive at x=0 after you multiply x throughout in order to solve the equation, as that would mean you had introduced an unnecessary common multiple into the equation..

I'm not at all clear what you are saying here. Certainly if you just take an arbitrary equation, say x²= 0, and multiply both sides by x, you can arrive at x= 0: x³= 0 so x= 0. You may be thinking of an equation in which you have good reason to multiply by x- say something like (x-1)/x= 0. In that case, even before you multiply by x, you know that x= 0 cannot be a solution.
It is well known that if you multiply both sides of an equation by any function of x, you may introduce "extraneous" roots: numbers that satisfy the new equation but not the original. The only way to be sure is to check the numbers in the original equation.

 

[truewt]

Alright thanks. Sorry for not visiting this thread after so long.