Extraneous solutions of radical equations (example 2) | High School Math | Khan Academy

We're asked which value for D we see D in this equation here makes x = -3 an extraneous solution for this radical equation. √(3x + 25) is equal to D + 2x, and I encourage you to pause the video and try to think about it on your own before we work through it together.

All right, now let's work through this together. So the first thing to just remind ourselves is what is an extraneous solution? Well, that's a solution that we get, or we think we get, but it's really just a byproduct of how we solved it. But it isn't going to be an actual solution of our original equation.

Now, how do these extraneous solutions pop up? Well, it pops up when you take the square of both sides. So for this equation right over here, to get rid of the radical, I'd want to square both sides of it. If I square both sides, the left-hand side will become 3x + 25, and the right-hand side, if I square this, is going to be D + 4DX + x^2. So that's just squaring both sides of this.

But notice, there's actually a different equation than this one that if you squared both sides you would also get this. What is that different equation? Well, the different equation is if you took the negative of one of these sides. So for example, if you had, if you started the original equation, the negative square root of 3x + 25 is equal to D + 2x.

You square both sides of this, you still get this purple equation. Because you square a negative, you get a positive. So both of them, when you square both sides, get us over here. And so when you solve this purple equation, this is going to be a quadratic right over here. You just rearrange it a little bit, you get it into standard quadratic form, you'll get two solutions.

It turns out one of the solutions is going to be for this yellow equation, and one of the solutions is going to be for the purple equation. And if the one, the solution that is for the purple equation is going to be an extraneous solution for the yellow equation, it's actually not going to be a solution for the yellow equation.

So when they say which value for D makes x = -3 an extraneous solution for this yellow equation, that's the same thing as saying what value of D makes x = -3 a solution for this. So a solution, a solution for this, if it's a solution for this, it's going to be an extraneous solution for that because these are two different equations.

We have this, we're taking the negative of just one side of this equation to get this one. If you took the negative of both sides, it becomes the same thing because you can multiply both sides of an equation times a negative value. So, a solution for this, which is equivalently a solution, which is equivalent to a solution, if I, instead of putting the negative on the left-hand side, if I multiplied the right-hand side by the negative.

But anyway, let's think about which value for D makes x = -3 a solution for this. Well, let's substitute x = -3 here, and then we just have to solve for D. If x = -3, then the square root of 3 * -3 is -9 + 25 is equal to D + 2 * -3 is -6. So, D = -6.

Now, we can square both sides. Actually, let's do it this way. I don't want to square both sides because we lose some information. This is going to be the negative square root of 9 + 25, which is 16, is equal to D - 6. So this is going to be equal to the negative. This is going to be equal to -4. The principal root of 16 is 4. We have the negative out front, which is equal to D - 6.

Then, adding six to both sides, you get 2 is equal to D. So, if D is equal to 2 here, if D is equal to 2, then a solution to this purple equation is going to be x = -3.

So that would be an extraneous solution because if x = -3 satisfies this over here, it's definitely going to satisfy this over here, but it's not going to satisfy this up here. And you can verify this: if this is equal to 2, try out x = -3. You're going to get on the left-hand side, you're going to get 16, and on the right-hand side, you're going to get 2 - 6, which is equal to -4.

So this does not work out. x = -3 is not a solution to this, but it is a solution for this, and it is a solution to this quadratic right over here. So, d = 2 makes x = -3 an extraneous solution for this equation.