Consider the equation x^3=x.

How many values of x satisfy this equation? Many students will divide both sides by x to get

x^2=1

Some will then take the square root of both sides of this equation to get

x=1

Looks like one solution. However, it turns out that the original equation has 3 solutions.

x^3=x.

Subtract x from both sides:

x^3-x=0

Factor x from both terms:

x(x^2-1)=0

Factor again:

x(x-1)(x+1)=0

Now we use the fact that terms can only multiply to zero if some of them equal zero to get the three solutions:

x=0, x-1=0, and x+1=0, so that

x=0, x=1, and x=-1 are the three solutions.

So

In the second step, where we took the square root of both sides, we also lost a solution because taking the square root only includes the positive square root. There is a positive and a negative square root.

To sum up, when dividing by a variable, we could potentially lose the x=0 solution. To avoid this issue, there are a couple of approaches. One is to

Similarly for avoiding the square root issue, we could factor, or we could remember to write a +/- sign to include both the positive and negative roots.

One other note is that, going the other direction,

Multiplying by 0 can potentially introduce false (aka extraneous) x=0 solutions, and squaring an equation can also similarly introduce a false solution.

For example, if we multiply both sides of

x=1 by x, we get

x^2=x, which now has 2 solutions, x=0 and x=1. We introduced the extraneous x=0 solution by multiplying by x.

Similarly, if x = 1, and we square both sides, we get that x^2=1, which has 2 solutions: x=1 and x=-1. We introduced the false extraneous negative solution by squaring.

Remember to plug the final solutions back in to the original equation to make sure no extraneous solutions were introduced.

The final summary:

If you found this helpful, remember to like or share!

How many values of x satisfy this equation? Many students will divide both sides by x to get

x^2=1

Some will then take the square root of both sides of this equation to get

x=1

Looks like one solution. However, it turns out that the original equation has 3 solutions.

**In both steps we lost a solution**. To see this, let's resolvex^3=x.

Subtract x from both sides:

x^3-x=0

Factor x from both terms:

x(x^2-1)=0

Factor again:

x(x-1)(x+1)=0

Now we use the fact that terms can only multiply to zero if some of them equal zero to get the three solutions:

x=0, x-1=0, and x+1=0, so that

x=0, x=1, and x=-1 are the three solutions.

So

**what went wrong?**Well it turns out that in the the first step at the very beginning of the post, when we divided by x, we could potentially be dividing by zero, which is not possible. In doing so, we lost the x=0 solution by canceling out the factor of x.In the second step, where we took the square root of both sides, we also lost a solution because taking the square root only includes the positive square root. There is a positive and a negative square root.

To sum up, when dividing by a variable, we could potentially lose the x=0 solution. To avoid this issue, there are a couple of approaches. One is to

**factor out the variable**, like we did the second time around, rather than dividing it out and canceling. The other is to**remember to write x=0**as a potential solution any time we divide by x.Similarly for avoiding the square root issue, we could factor, or we could remember to write a +/- sign to include both the positive and negative roots.

One other note is that, going the other direction,

Multiplying by 0 can potentially introduce false (aka extraneous) x=0 solutions, and squaring an equation can also similarly introduce a false solution.

For example, if we multiply both sides of

x=1 by x, we get

x^2=x, which now has 2 solutions, x=0 and x=1. We introduced the extraneous x=0 solution by multiplying by x.

Similarly, if x = 1, and we square both sides, we get that x^2=1, which has 2 solutions: x=1 and x=-1. We introduced the false extraneous negative solution by squaring.

Remember to plug the final solutions back in to the original equation to make sure no extraneous solutions were introduced.

The final summary:

__Dividing__or__taking roots__can__drop solutions__.__Multiplying__or__taking exponents__can__introduce extraneous solutions__(so make sure to plug your solutions obtained to eliminate extraneous solutions).If you found this helpful, remember to like or share!