When and why extraneous solution happen

In other videos, we've already introduced the idea of an extraneous solution, where you go about solving an equation. You're given an original equation, and you do a bunch of algebraic steps. Then you solve it and you get some solutions. What we've seen, especially when we're dealing with radical equations and especially some rational equations, is that sometimes all of the solutions don't work for our original equation. The solutions that don't work are the ones that we call extraneous.

What we're going to do in this video is get a little bit deeper. We're going to look at some extraneous solutions, but really think deeply about why they show up. So let's just give an example, and this is one that hopefully you're familiar with, or at least the idea of it: you're familiar with when you're dealing with radical equations.

Let's say you're trying to solve (2x - 1 = \sqrt{8 - x}). I'm going to go quickly through it; if this is unfamiliar to you, I encourage you to review solving radical equations. But we're going to really study why we get an extraneous solution here.

So actually, why don't you pause this video, figure out what the solutions are? I'll give you a hint: one of them is going to be extraneous, and really think about why that extraneous solution shows up. Alright, now let's do it together.

Just to go through the steps that we've seen before for radical equations, you want to square both sides of this. So if you square the left-hand side, you'll get (4x - 4x). I should say, minus (4x + 1) is equal to you square the square root of (8 - x). You're going to get just (8 - x).

Now let's see: we can subtract (8) from both sides, and we can add (x) to both sides. So we just have (0) on the right-hand side, and on the left-hand side, we're going to get (4x^2 - 3x - 7 = 0). So we have an interesting quadratic here. You could use the quadratic formula, but it looks like this one might be factorable by grouping.

We just have to think of two numbers that, when we add them, so (a + b), need to be equal to (-3), and (a \times b) needs to be equal to (4 \times -7) which is equal to (-28). The obvious ones seem to be right in front of us: (4) and (-7).

Four plus negative seven is equal to negative three, and four times negative seven is equal to negative twenty-eight. So we can just break this negative three up into, or negative (3x) up into (4x) and a negative (7x). So let me do that.

So you have (4x^2), and then I'm just going to – and this is just factoring by grouping. Once again, if this is unfamiliar, I encourage you to review that on Khan Academy. So we're going to get plus (4x - 7x). Notice these two add up to negative (3x), so I'm not changing the equation really, or the value, or what it's trying to express, is equal to (0).

In factoring by grouping, we can group these two together, and then we can group these two together. For these first two, we can factor out a (4x) and so you're going to get a (4x \times (x + 1)), and in the second two, we can factor out a negative (7): (-7 \times (x + 1)). And all of that is going to be equal to zero.

Then we can factor out an (x + 1), and so we're going to get ((x + 1)(4x - 7) = 0). Once again, this is just to get us to a place where we can look at what the solutions are.

So we can see that either (x + 1 = 0) which happens when (x = -1), or (4x - 7 = 0) which happens when (4x = 7), which happens when (x = \frac{7}{4}). So we got these two solutions, and I promised you that one of them would be extraneous.

So let's try this first one, (x = -1), and remember when you solve a radical equation like this, you want to substitute back in the original equation to see whether they're extraneous or not.

So when you substitute (x = -1) back here, you're going to get (2(-1) - 1 = \sqrt{8 - (-1)}). On the left-hand side, you have (-2 - 1 = -3), and you get that equaling the square root of (9): the principal root of (9).

So you get this weird result that (-3 = 3), which we know is not the case, and that's why we know this is extraneous. But, as I promised, why did this happen?

The key is to realize that when you're doing algebraic manipulation, some operations are reversible and some are not. For example, if you say that (a + b = c) and you want to solve for (a) and you subtract (b) from both sides, you are going to get (a = c - b).

If you were to start off with (a = c - b), you could add (b) to both sides, and you could add (b) to both sides to get back to your original equation: (a + b = c).

So this adding and subtracting values from both sides is a reversible operation; you can do it in either direction. But when you're squaring things, that is not true. We did square right over here, did we? To go from that first step to this step, we squared.

What we'll see, or what you might already recognize, is that squaring is not reversible. If I know that (a = b), and then I were to square it, we do know that (a^2 = b^2). But if we go the other way around, if we know that (a^2 = b^2), do we know that it's always the case that (a = b)?

I'll give you an example that shows that that is not always true. We know that ((-2)^2 = 2^2), but (-2) is not equal to (2). So that's what we did here. We squared at this step and it's not reversible.

We were able to deduce a bunch of stuff, and then we got all the way to the point of saying okay, (x = -1) or (x = \frac{7}{4}). But it's not always the case that when (x = -1) that we can reverse all the way back to the original equation (2x - 1 = \sqrt{8 - x}).

The two places where this is important when you're dealing with algebra—the two obvious places—are when you're dealing with radical equations because you're doing that squaring step, and that squaring step is not reversible.

The other case is when you are dealing with multiplying both sides by a variable. To understand that, if we know that (a = b), but then we were to multiply both sides of this equation by (x), we'll get (xa = xb). That could be true, but is the other way around necessarily true?

If we know that (xa = xb), is it always the case that (a = b)? I'll give you an example where that isn't the case. What happens when (x = 0)? We know that (0 \times 3 = 0 \times 4), but it is not the case that (3 = 4).

This is another situation where multiplying both sides of an equation by a variable—you can go in that direction, but it's not necessarily reversible. You can't necessarily go back in that direction.

That's why, when you solve an equation, oftentimes a rational equation by multiplying both sides of that equation by a variable, and you get a result, it's not necessarily true that you could take that result and reverse everything and get back to the original equation.

And that's why you have to check your extraneous solutions, especially for radical equations, and especially when you do an algebraic operation like multiplying both sides of an equation by a variable or by an expression that involves a variable.