Video: Extraneous Solutions with Radicals

Video Transcript

The natural way to solve the square root of 𝑥 plus nine minus two 𝑥 equals three introduces an extraneous solution. What is that extraneous solution?

So what is the natural way to solve this equation? First, we just write down the equation we want to solve. Here, I’ve added two 𝑥 to both sides to get the radical isolated on the left-hand side. I can square both sides so the square root of 𝑥 plus nine becomes just 𝑥 plus nine, and two 𝑥 plus three becomes two 𝑥 plus three squared. Expanding the brackets on the right-hand side, I can see that 𝑥 plus nine is equal to four 𝑥 squared plus 12𝑥 plus nine.

We can swap the two sides so now we have four 𝑥 squared plus 12𝑥 plus nine equals 𝑥 plus nine. Subtracting nine from both sides, we get four 𝑥 squared plus 12𝑥 is equal to 𝑥. Subtracting 𝑥 from both sides we get four 𝑥 squared plus 11𝑥 equals zero. Factorising, we get that 𝑥 times four 𝑥 plus 11 is equal to zero. We have the product of two things being zero, so one of them must be zero, where the 𝑥 is zero or four 𝑥 plus 11 is zero. And in the latter case that means that 𝑥 must be negative 11 over four.

Okay. So these two values of 𝑥 are clearly two solutions to the last equation that we got. The question is are they also solutions to the first equation, the equation we started with, square root of 𝑥 plus nine minus two 𝑥 equals three.

Let’s first check 𝑥 equals zero. Substituting 𝑥 for zero, we get the square root of zero plus nine minus two times zero is equal to three which means that the square root of nine minus zero equals three, and that the square root of nine is equal to three, which of course it is. So 𝑥 equals zero is a solution to our original equation; there’s no problem here.

How about 𝑥 equals negative 11 over four? It’s the same process. We substitute in negative 11 over four. Simplifying somewhat, we get that the square root of 25 over four plus 11 over two should be equal to three. And hence that the square root of 25 over four is equal to negative five over two. It turns out that the square root of 25 over four is just five over two. In the process of checking that 𝑥 equals negative 11 over four was a solution to our original equation, we’ve got something which patently isn’t true. We’ve got five over two is equal to negative five over two. And so we see that 𝑥 equals negative 11 over four can’t be a solution to our original equation.

So what went wrong? The answer is that 𝑥 equals negative 11 over four is an extraneous solution that we’ve introduced. As such, it is also the answer to our question. You can check that this value satisfies the equation 𝑥 plus nine equals two 𝑥 plus three all squared. As a result, it also satisfies all the equations that come after that one, including the last equation which we read the solutions from. However, you can check that it does not satisfy the equation above, the square root of 𝑥 plus nine equals two 𝑥 plus three. Substituting this value into the equation gives five over two equals negative five over two which is of course not true. When you square both sides of this equation, you get something which is true five over two squared does equal negative five over two squared.

It was the squaring step where we introduced the extraneous solution. The good news is that squaring both sides of an equation only ever adds solutions. You can never lose a solution by squaring both sides. So if you square both sides in the process of solving an equation, all you have to do is check that all the solutions to the equation you get at the end also satisfy the equation that you started with. If you have a value which satisfies the last equation but not the first equation, then that’s an extraneous solution, and you need to remove it.