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CCSS.Math: ,

- [Instructor] 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. And so the key of this video is why do extraneous solutions even occur? And it all is due to the
notion of reversibility. There's 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. And I'll show you those two operations. One is squaring and the other
is multiplying both sides by a variable expression. So let's just 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 a example that shows that this is not always the case? Actually, pause the video,
try to think about it. Well, negative two squared is indeed equal to two squared, but negative two is not equal to two. 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, like I said,
that will make us confused, looks like an x. Multiply both sides by 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. xa is going to be equal to xb. But the other, the reverse,
isn't always the case. If xa is equal to xb, is it always the case
that a is equal to b? Well, the simple answer is no, and I always encourage
you, pause this video and see if you can find an example where this doesn't work. Well, if a was two and b is three and the variable x just happened
to take on the value zero. So we know that zero times two is indeed equal to zero times three, but two is not equal to three. Now, how does all this connect to the extraneous solutions you've seen when you were solving radical equations or when you were solving some rational, or equations with rational
expressions on both sides? Well, let's look at an example. Let's solve a radical equation. If I wanted to solve the equation the square root of five x minus four is equal to x minus two, a typical first step is, hey,
let's get rid of this radical by squaring both sides. So I'm going to square both sides, and then I'm going to
get five x minus four is equal to x squared
minus four x plus four. 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. And let's see, we can subtract
five x from both sides. We can add four to both sides. I'm just trying to get a
zero on the left hand side. And so I'm going to be left with zero is equal to x squared
minus nine x plus eight, or zero is equal to x minus
eight times x minus one, or we could say that x
minus one is equal to zero, or x minus eight is equal to zero. We get x equals one or x equals eight. So let's test these solutions. If x equals eight, we would get, and I'll color code it a little bit, for x equals eight, if I test
it in the original equation, I get the square root
of 36 is equal to six, which absolutely true, so that one works. But what about x equals one? I get the square root of five
times one minus four is one is equal to one minus two,
which is equal to negative one. 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 one, even though you got there with
legitimate algebraic steps. And the reason that is true is, actually, pause this video, look back. For which of these steps
does x equal one still work and what step does it not work? Well, you'll see that x equals one works for all of these equations
below this purple line. It just doesn't work for the square root of five minus four x is equal to x minus two. In fact, you could start with x minus one 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 one. And we could do the same
thing with a rational, or an 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 one is equal to one over x minus one, the first thing I might wanna do is multiply both sides by x minus one. So multiply by x minus one. Now, notice I'm multiplying both sides by a variable expression, so we have to be a little bit contentious now, but if I'd multiply both
sides by x minus one, I'm going to get x
squared is equal to one, or I could say that x equals one or x is equal to negative one. Well, we could test these. For x equals one, if I go up here, I'm dividing by zero on both sides. So this is an extraneous solution. They key here is that
we multiplied both sides by a variable expression. In this case, we multiplied
both sides by x minus one. 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 zero times two
is equal to zero times three does not mean that two is equal to three. It's completely analogous because we multiplied
by a variable expression that actually takes on the value zero when x is equal to one. 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 could add or subtract anything from both sides of an equation, and that's always going to be reversible. And so that's not going to
lead to extraneous solutions. You could 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.