- [Instructor] In other videos,
we've already introduced the idea of an extraneous solution
where you go about solving an equation.
So you're given an original equation,
and you do a bunch of algebraic steps,
and then you solve it, and you get some solutions.
And what we've seen, especially when
we're dealing with radical equations
and especially some rational equations,
sometimes all of the solutions don't work for
our original equation.
And so the solutions that don't work,
that don't work, these are the ones
that we call extraneous.
What we're going to do in this video is
get a little bit deeper.
We're gonna 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 two x minus one is equal
to the square root of eight minus x.
And I'm gonna go quickly through it.
If this is unfamiliar to you, I encourage
you to review solving radical equations.
But we're gonna 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's going to be extraneous.
And really think about why
that extraneous solution shows up.
All right, now let's do it together
and just to go through the steps that
we've seen before for radical equations.
You wanna square both sides of this.
So if you square the left-hand side,
you'll get four x squared, I should say,
minus four x plus one is equal to,
you square the square root of eight minus x.
You're going to get just eight minus x.
And now let's see.
We can subtract eight from both sides,
and we can add x to both sides,
so we just have zero on the right-hand side.
And on the left-hand side we're going
to get four x squared minus three x minus seven is equal
to zero, 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 plus b needs
to be equal to negative three,
and a times b needs to be equal
to four times negative seven,
needs to be equal to negative 28.
And the obvious ones seem to be right in front of us,
four and negative seven.
Four plus negative seven is equal to negative three.
Four times negative seven is equal to negative 28.
So we can just break this negative three up
into, or negative three x up into a four x
and a negative seven x.
So let me do that.
So you have four x squared, 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 four x minus seven x,
and notice these two add up to negative three x.
So I'm not changing the equation, really,
or the value, or what it's trying to express,
is equal to zero.
And in factoring by grouping we can group
these two together, and then we can group
these two together.
And for these first two, we can factor
out a four x, and so you're going
to get a four x times x plus one.
And in these second two we can factor
out a negative seven, minus seven times x plus one,
and all of that is going to be equal to zero.
And then we can factor out an x plus one.
And so we're going to get x plus one
times four x minus seven.
So times four x minus seven is equal to zero.
And 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 plus one is going
to be equal to zero, which happens
when x is equal to negative one,
or four x minus seven is equal to zero,
which happens when four x is equal to seven,
which happens when x is equal to 7/4.
So we've got these two solutions and I promised you
that one of them would be extraneous.
So let's try this first one, x equals negative one.
And remember, when you solve a radical equation like this
you wanna substitute back into the original equation
to see whether they're extraneous or not.
So when you substitute x equals negative one back here,
you're going to get two times negative one
minus one is going to be equal to the square root
of eight minus negative one,
and so on the left-hand side
you have negative two minus one.
That is negative three.
And you get that equaling the square root of nine,
the principal root of nine, and so you get this weird result
that negative three is equal to three,
which we know is not the case, and that's why
we know this is extraneous.
But as I promised, why did this happen?
And 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 plus b is equal
to c, and you wanna solve for a,
and you subtract b from both sides,
you are going to get a is equal to c minus b.
And if you were to start off with a is equal
to c minus b, you could add b
to both sides.
You could add b to both sides to get back
to your original equation,
to get a plus b is equal to c.
So this adding and subtracting values
from both sides, that is a reversible operation.
You can do it in either direction.
But when you're squaring things,
that is not true, and we did square right over here.
To go from that first step to this step, we squared.
And what we'll see, or what you might already recognize is
that squaring is not reversible.
If I know that a is equal to to b,
and then I were to square it, we do know
that a squared is equal to to b squared.
But if we go the other way around,
if we know that a squared is equal to b squared,
so if we know that a squared is equal to b squared
do we know that it's always the case
that a is equal to b?
Well I'll give you an example that shows
that that is not always true.
We know that negative two squared is equal
to two squared, but negative two is not equal to two.
So that's what we did here.
We squared at this step, and it's not reversible,
and we were able to do a bunch of stuff,
and we got all the way to the point
that saying, okay, x is equal to negative one,
or x is equal to 7/4, but it's not always the case
that when x equals negative one
that we can reverse all the way back
to this original equation and that two x minus one is equal
to the square root of eight minus x.
And 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, 'cause 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.
And to understand that, if we know
that a is equal to to b, but then
we were to multiply both sides of this equation times x.
And so we'll get xa is equal to xb.
That could be true, but is the other way
around necessarily true?
If we know that xa is equal to xb,
is it always the case that a is equal to b?
Well, I'll give you an example where
that isn't the case.
What happens when x is equal to zero?
We know that zero times three is equal to zero times four,
but it is not the case that three is equal to four.
And so 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.
And 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 your 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.