Extraneous solutions | Equations | Algebra 2 | Khan Academy

In this video, we're going to talk about extraneous solutions. If you've never heard the term before, I encourage you to review some videos on Khan Academy on extraneous solutions. But just as a bit of a refresher, it's the idea that you do a bunch of legitimate algebraic operations, you get a solution or some solutions at the end, but then when you test it in the original equation, it doesn't satisfy the original equation.

So, the key of this video is: why do extraneous solutions even occur? It all is due to the notion of reversibility. There are certain operations in algebra that you can do in one direction, but you can't— and it'll always be true in one direction, but it isn't always true in the other direction. I’ll show you those two operations: one is squaring, and the other is multiplying both sides by a variable expression.

Let's see the example of squaring, and then we're going to see it in an actual scenario where you're dealing with an extraneous solution. So we know, for example, that if a is equal to b, I could square both sides, and then a squared is going to be equal to b squared. But the other way is not true. For example, if a squared is equal to b squared, it is not always the case that a is equal to b.

What's an example that shows that this is not always the case? Actually, pause the video, try to think about it. Well, negative 2 squared is indeed equal to 2 squared, but negative 2 is not equal to 2. So this shows that you can square both sides of an equation and deduce something that is true, but the other way around is not necessarily going to be true.

Another non-reversible operation sometimes is multiplying both sides by a variable expression. So multiplying both sides— actually, that gets confused— looks like an x. Multiply both sides by a variable— I’ll just write ‘variable’, but it could be a variable expression as well. For example, we know that if a is equal to b, that if we multiply both sides by a variable, that's still going to be true. x times a is going to be equal to x times b.

But the reverse isn’t always the case. If x times a is equal to x times b, is it always the case that a is equal to b? Well, the simple answer is no. I always encourage you to pause this video and see if you can find out an example where this doesn't work. Well, what if a was 2 and b is 3, and the variable x just happened to take on the value 0?

So, we know that 0 times 2 is indeed equal to 0 times 3, but 2 is not equal to 3. Now, how does all this connect to the extraneous solutions you've seen when you're solving radical equations or when you're solving some rational equations with rational expressions on both sides?

Let's look at an example. Let's solve a radical equation. If I wanted to solve the equation, the square root of 5x minus 4 is equal to x minus 2. A typical first step is, "Hey, let's get rid of this radical by squaring both sides." So I am going to square both sides, and then I'm going to get 5x minus 4 is equal to x squared minus 4x plus 4.

Once again, if this looks completely unfamiliar to you, we go into much more depth in other videos where we introduce the idea of radical equations. Let's see. We can subtract 5x from both sides; we can add 4 to both sides. I'm just trying to get a 0 on the left-hand side. So I'm going to be left with 0 is equal to x squared minus 9x plus 8, or 0 is equal to (x minus 8)(x minus 1).

Or we could say that x minus 1 is equal to 0 or x minus 8 is equal to 0. We get x equals 1 or x equals 8. So let's test these solutions. If x equals 8, we would get— and I'll color code it a little bit for x equals 8— if I test it in the original equation, I get the square root of 36 is equal to 6, which is absolutely true. So that one works.

But what about x equals 1? I get the square root of 5 times 1 minus 4 is equal to 1, which is equal to 1 minus 2, which is equal to negative 1. That did not work. This right over here is an extraneous solution.

If someone said, "What are all the x values that satisfy this equation?" you would not say x equals 1, even though you got there with legitimate algebraic steps. The reason that is true is— actually pause this video, look back for which of these steps does x equal 1 still work, and what step does it not work. Well, you'll see that x equals 1 works for all of these equations below this purple line; it just doesn't work for the square root of 5 minus 4 is equal to x minus 2.

In fact, you can start with x minus 1 and then you could deduce all the way up to this line here, but the issue here is that squaring is not a reversible operation. This is analogous to saying, "Hey, we know that a squared is equal to b squared." We know that this is equal to this, but then that doesn’t mean that a is necessarily equal to b for x equals 1.

We could do the same thing with a rational equation that deals with rational expressions. So, for example, we might have to deal with— and let me make sure I have some space here— if I had to solve x squared over x minus 1 is equal to 1 over x minus 1. The first thing I might want to do is multiply both sides by x minus 1. So multiply by x minus 1.

Now notice I'm multiplying both sides by a variable expression, so we have to be a little bit conscientious now. But if I multiply both sides by x minus 1, I'm going to get x squared is equal to 1 or I could say that x equals 1 or x is equal to negative 1. But we could test these for x equals 1— if I go up here, I'm dividing by 0 on both sides, so this is an extraneous solution.

The key here is that we multiplied both sides by a variable expression. In this case, we multiplied both sides by x minus 1. You can do that; you can multiply both sides by a variable expression, and it is a legitimate algebraic operation. It's completely analogous to what we saw right over here. Just because 0 times 2 is equal to 0 times 3 does not mean that 2 is equal to 3.

It's completely analogous because we multiplied by a variable expression that actually takes on the value 0 when x is equal to 1. So the big takeaway here is hopefully you understand why extraneous solutions happen a little bit more. When you square, when you multiply both sides by a variable expression— completely legitimate as long as you do it properly. But it's not always the case that the reverse is true.

You can add or subtract anything from both sides of an equation, and that's always going to be reversible. So that's not going to lead to extraneous solutions. You can multiply or divide by a non-zero constant value— that's also not going to lead to anything shady. But if you're squaring both sides or multiplying both sides by a variable expression, you should be a little bit careful.