STATEMENT: At school, we are told:

"whatever we do to one side of an equation, you must (and can) do to the other side."

Example 1: Let us start off with a simple equation, such as 2x+3=6

STEP 1: We subtract 3 from both sides

2x=3

STEP 2: Divide both sides by 2

x=3/2

Everything is correct.

Now look at Example 2: x^2=(x+1)^2

The "correct" way to do this is to multiply the brackets, collect like terms and solve.

But if the statement at the top of this post is correct, then watch what happens if we take the following steps:

STEP 1: Take the square root of both sides of the equation

x=-(x+1) OR x=x+1

which means x=-1/2 OR 0=1!

Clearly the only correct answer is x=-1/2

So my question is: is it correct to say:

"whatever we do to one side of an equation, you must (and can) do to the other side."

It's true but it's incomplete. That statement has no mention of choices. At that point you branch and throw out the contradictions.

This makes no sense. What do you mean by "a solution" to f(x)? The roots?? It's not clear.

Simplifying: f(x) = x^2/3 + 2x/3 which does not contain the parameter "a" so "a" has no meaning.

So I am at a loss.

Please help me,. (prove : a is a only solution of function f(x)=(x)^2÷3 + x + (-x)1÷3 on -inf;-1] then a^3 +4x^2 -x=0

Things get tricky when square rooting both sides. You have to be very careful in your use of +/- or you can get a nonsensical statement. You have to pay attention as to the interval so see whether the LS and RS are + or - in that interval. We want them both to be positive.

Note that in the interval -1<= x <= 0:

sqrt[x^2] = -x and not x

sqrt[(x+1)^2] = x+1 and not -(x+1)

This ensures that both LS and RS are positive and leads to x = -1/2

sqrt[x^2] can never equal sqrt[(x+1)^2] when x >0 or x<-1 so x=x+1 can never happen

So for the 4 basic operations your statement is true. It is also true for the sqrt operation but you have to be careful in choosing + or - and which ones work in the interval your root is in. By equating x = x+1 you are trying to make a positive number = to a negative number (in that interval) and you have pointed out what happens when you do that.

Yes correct where as it is necessary to keep the equation balance which would be possible in this manner also.