Solving for x, but only getting half

[Calpalned]

Homework Statement

Solve for x. x^2 - 2x = 0

Homework Equations

n/a

The Attempt at a Solution

If I solve this equation by factoring out the x first, as in x(x-2) = 0, I get that x = 2 or x =0

However, if I add 2x, as in x^2 = 2x, then I divide by x, leaving x = 2
My second method only has one answer. What happened to x = 0?
When I solve equations like this, what should I do to guarantee that I get all of the solutions without using a graphing calculator?

Thank you

 

[Fightfish]

However, if I add 2x, as in x^2 = 2x, then I divide by x, leaving x = 2
My second method only has one answer. What happened to x = 0?

You cannot divide by x if it is zero. So your method assumes that x is not zero already, and obviously then you can't get it as a solution! In solving such problems, never divide by a quantity unless (1) you are sure it is nonzero or (2) you split out a separate case for when it is zero.

 

[Shinaolord]

Maybe $x^2-2x = x(x-2)$ will help you? There are other ways as well. Such as $x^2=2x$. Note that for the 2nd equation 0 is a solution ($0^2 = 2*0$). Then we can proceed to find non-zero solutions.

 

[Calpalned]

Thanks everyone!

 

[Mark44]

Maybe $x^2-2x = x(x-2)$ will help you? There are other ways as well. Such as $x^2=2x$.

Writing the equation as x² = 2x is pretty much a step backward, as it isn't any help. The only two ways I can think of to solve x² - 2x = 0 are 1) factoring, and 2) quadratic formula. (In case someone suggests completing the square, that's already done for you in the quadratic formula.)

For this equation, it is much simpler to just factor the terms on the left side.

As already mentioned, any temptation to divide through by x should be resisted, due to the possibility of losing solutions.

Note that for the 2nd equation 0 is a solution ($0^2 = 2*0$). Then we can proceed to find non-zero solutions.

 

[Fredrik]

$x^2=2x$ implies that either $x=0$ or $x=2$ (because if $x\neq 0$, we can divide by $x$ to get $x=2$).

The above doesn't prove that the solutions are 0 and 2. It just proves that no number other than 0 or 2 can be a solution. It is however easy to verify that both 0 and 2 are solutions: Since $0^2=0=2\cdot 0$, we know that $0$ is a solution. Since $2^2=2\cdot 2$, we know that $2$ is a solution.

This is how I think when I solve these problems. I like this method better than both factorization and the formula for solutions.

An alternative to explicitly verifying that 2 is a solution is to note that if $x\neq 0$, the statements $x^2-2x=0$, $x^2=2x$ and $x=2$ are all equivalent.

In case it's not perfectly clear what's going on in the first step: We already knew without looking at the equation that either $x=0$ or $x\neq 0$. So we consider these two possibilities separately. If $x=0$, the equation tells us nothing. (OK, it tells us that $0=0$, but we already knew that). If $x\neq 0$, then the equation tells us that $x=2$. We have now ruled out all non-zero solutions other than $2$.