Solving square-root equations: no solution | Mathematics III | High School Math | Khan Academy

Let’s say that we have the radical equation: the square root of 3x minus seven plus the square root of 2x minus one is equal to zero.

I encourage you to pause the video and see if you can solve for X before we work through it together.

Alright, so one thing we could do is we could try to isolate each of the radicals on either side of the equation. So let’s subtract two; let’s subtract this one from both sides so I can get it onto the right-hand side—or a version of it on the right-hand side.

So, I’m subtracting it from the left-hand side and from the right-hand side, and so this is going to get us that. It is going to get us on the left-hand side; I just have square root. These cancel out, so I’m just left with the square root of 3x minus seven is going to be equal to this: the negative of the square root of 2x minus one.

So now we can square both sides, and we always have to be careful when we’re doing that because whether we’re squaring the positive or the negative square root here, we’re going to get the same value.

So the solution we might get might be the version when we're solving for the positive square root, not when we take the negative of it.

We have to test our solutions at the end to make sure that they’re actually valid for our original equation. But if you square both sides, on the left-hand side we are going to get 3x minus seven, and on the right-hand side, a negative square is just positive, and the square root of 2x minus one squared is going to be 2x minus one.

Now, see, we could subtract 2x from both sides to get all of our X’s on one side, so I’m trying to get rid of this, and we can add 7 to both sides because I’m trying to get rid of the negative 7.

So, add 7 to both sides and we are going to get: 3x minus 2x is X, is equal to negative 1 plus 7, X is equal to 6.

Now, let’s verify that this actually works. So if we look at our original equation, the square root of 3 times six minus seven, minus seven needs to be equal to zero.

So does this actually work out? Three times six minus seven, so this is going to be the square root of eleven plus the square root of eleven needs to be equal to zero, which clearly is not going to be equal to zero.

This is two square roots of 11, which does not equal zero, so this does not work. And you might say: “Wait, how did this happen? I did all of this nice neat algebra, I didn’t make any mistakes, but I got something that doesn’t work.”

Well, this right here is an extraneous solution. Why is it an extraneous solution? Because it’s actually the solution to the equation: the square root of three X minus seven minus the square root of two X minus one is equal to zero.

And you might say: “Well, if it’s a solution to that, if it’s the solution to this thing right over here, how did I get the answer while I’m trying to do algebraic steps there?”

Well, the key is when we added—when we took this onto the right-hand side and squared it, well, it all boiled down to this. Regardless of what starting point you started with, if you did the exact same thing, you would’ve gotten to that same point right over here.

So the solution to this ended up being the solution to this starting point versus the one that we originally started with.

Interestingly, this one has no solutions, and it actually would be fun to think about why it has no solutions. We’ve shown, to a certain degree, the only solution you got by taking reasonable algebraic steps is an extraneous one—it’s a solution to a different equation that has a common intermediate step.

But it’s also fun to think about why this right over here is impossible.