Extraneous solutions of radical equations | Mathematics III | High School Math | Khan Academy

Let's say we have the radical equation (2x - 1 = \sqrt{8 - x}).

So we already have the radical isolated on one side of the equation. We might say, "Well, let's just get rid of the radical; let's square both sides of this equation."

So we might say that this is the same thing as ( (2x - 1)^2 = (8 - x) ). Then we would get, let's see, ( (2x - 1)^2 = 4x^2 - 4x + 1 ) is equal to ( 8 - x).

Now we have to be very, very careful here. We might feel, "Okay, we did legitimate operations; we did the same thing to both sides," that these are equivalent equations, but they aren't quite equivalent.

Because when you're squaring something, one way to think about it is, when you're squaring it, you're losing information. For example, this would be true even if the original equation were (2x - 1 = -\sqrt{8 - x}).

Because if you squared both sides of this, you would also get that right over there, because the negative squared would be equal to a positive. So, when we're finding a solution to this, we need to test our solution to make sure it's truly the solution to this first yellow equation here and not the solution to this up here.

If it's a solution to the right-hand side or not the yellow one, then we would call that an extraneous solution.

So let's see if we can solve this. Let's write this as kind of a standard quadratic.

Let's subtract 8 from both sides to get rid of this 8 over here, and let's add (x) to both sides, so (+x).

We are going to get (4x^2 - 3x - 7 = 0).

Now, let's see; we would want to factor this right over here, and let's see, maybe I could do this by... well, I'll just use the quadratic formula here.

So the solutions are going to be (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}).

So (x) is going to be equal to (-(-3) + \sqrt{(-3)^2 - 4(4)(-7)}) over (2(4)).

Thus, (x) is equal to (\frac{3 \pm \sqrt{9 + 112}}{8}).

Let me make sure I'm doing this right. So (9 + 112 = 121).

That worked out nicely, so we have (3 \pm \sqrt{121}) all over (8).

Well, that is equal to (3 \pm 11) all over (8).

So that is equal to, if we add (11), that is (14/8), or if we subtract (11), (3 - 11 = -8), negative 8 divided by (8) is (-1).

So we have to think about, you might say, "Okay, I found two solutions to the radical equation."

But remember, one of these might be solutions to this alternate radical equation that got lost when we squared both sides.

We have to make sure that they're legitimate or maybe one of these is an extraneous solution.

In fact, one is very likely the solution to this radical equation, which wasn't our original goal.

So let's see. Let's try out (x = -1).

If (x = -1), we would have (2(-1) - 1 = \sqrt{8 - (-1)}).

So that would be (-2 - 1 = \sqrt{9}).

And so we'd have (-3 = \sqrt{9}).

The principal number, right, this is the positive square root.

This is not true, so this right over here is an extraneous solution.

It is a solution to this one right over here, because notice for that one, if you substitute (2(-1) - 1 = -(\sqrt{8 - (-1)})), so this is (-3 = -3).

So it checks out for this one.

So this one right over here is the extraneous solution, while this one right over here is going to be the actual solution for our original equation.

And you can test it out on your own; in fact, I encourage you to do so.