# Video: Finding the Solution Set of Radical Equations

What is the solution set of the equation 𝑥 + √(𝑥² + 20𝑥 + 100) = 5?

03:04

### Video Transcript

What is the solution set of the equation 𝑥 plus the square root of 𝑥 squared plus 20𝑥 plus 100 equals five?

The first step is just to write down the equation again. We can see that the main difficulty in solving this equation is going to be this square root term. We don’t really want to square root. And so at some point, we’re going to have to square to get rid of it. But if we square both sides now, on the left-hand side, we’ll get a cross term of two times 𝑥 times the square root of 𝑥 squared plus 20𝑥 plus 100. So we’ll still have a square root. So we haven’t really succeeded in getting rid of the square root. Our first step should be something else instead.

What we do first is to subtract 𝑥 from both sides. So the square root is isolated on the left-hand side. And now when we square both sides, we don’t get this cross term. We can expand the right-hand side. And we notice that we have an 𝑥 squared term on the both sides that cancel. Alternatively, you can think of this as subtracting 𝑥 squared from both sides. We then have a linear equation which we can solve in the normal way. Adding 10𝑥 to both sides, so we have an 𝑥 term on the left-hand side only. Subtracting by 100 on both sides to isolate this 𝑥 term. And then dividing by 30 on both sides. And simplifying the fraction we get on the right-hand side to find that 𝑥 is negative five over two.

Having got this answer, it’s really important that we check that it satisfies the original equation. That is, 𝑥 plus the square root of 𝑥 squared plus 20𝑥 plus 100 equals five. This isn’t just to make sure that we haven’t made any silly errors in our algebra. Although of course, it will catch those as well. It is that in the process of squaring we can introduce extraneous solutions to our problem. That is, values of 𝑥 which satisfy the squared equation but not the equation in the line before.

So let’s check that our candidate solution, 𝑥 equals negative five over two, is actually a solution to our original equation. We do this by substituting the value of negative five over two into the equation. So we’re hoping that negative five over two plus the square root of negative five over two squared plus 20 times negative five over two plus 100 is equal to five. And either by evaluating the left-hand side by hand or by using a calculator, we can see that this equation is indeed true.

And so we can see that our candidate solution, 𝑥 equals negative five over two, is indeed a solution to our original equation. We’re almost done. We found the solution. But we’re asked for the solution set. So we have to take our solution and put it in a set.

The solution set of the equation, 𝑥 plus the square root of 𝑥 squared plus 20𝑥 plus 100 equals five, is the set containing negative five over two.