### Video Transcript

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

First, letβs ask what is an extraneous solution? An extraneous solution is a solution that emerges from the process of solving the problem, that is not actually a valid solution to the problem. Before we find any extraneous solutions, we need to go ahead and solve this equation.

The first thing that I noticed is that Iβm going to need to square the square root of π₯ plus nine. In order to do that, Iβll have to move this negative two π₯ to the other side of the equation. We can do that by adding two π₯ to both sides of the equation. The two π₯s on the left side cancel out leaving us with the square root of π₯ plus nine equals two π₯ plus three. Now I can go ahead and square both sides of this equation. On the left side, the square root squared will cancel out leaving us with π₯ plus nine. On the right side, two π₯ plus three squared would look like two π₯ plus three times two π₯ plus three.

I want to go ahead and get rid of these parentheses which means Iβll need to multiply all the terms. Two π₯ times two π₯ equals four π₯ squared, two π₯ times three equals six π₯, three times two π₯ equals six π₯, and three times three equals nine. From there, I will bring down my four π₯ squared. I can combine six π₯ and six π₯; together, they would be 12π₯ and then bring down the nine. The left side of our equation hasnβt changed.

But to go from here to a solution, I wanna move everything to the same side of the equation. This means Iβll move the nine π₯ by subtracting nine π₯ from both sides. Nine minus nine equals zero; they cancel out. Weβre left with π₯ on the left. But if you notice on the right-hand side, itβs also positive nine minus nine. That will also drop out. Now we have π₯ equals four π₯ squared plus 12π₯. So we want to move our π₯ from the left side to the right side by subtracting π₯ on both sides. π₯ minus π₯ cancels out; they equal zero. We can bring down the four π₯ squared. 12π₯ minus π₯ equals 11π₯.

Weβre getting closer to our two solutions. But now we need to factor what we have here. Four π₯ squared and 11π₯ both have a factor of π₯. If we pull out the factor of π₯, we have π₯ times four π₯ plus 11 equals zero. We then set each of these factors equal to zero. The first solution would just be π₯ equals zero. Our second solution, weβre going to have to solve for because right now, we have four π₯ plus 11 equals zero. I solve for π₯ by subtracting 11 from both sides. Positive 11 minus 11 cancels out. Now we have four π₯ equals negative 11. We can divide both sides by four to isolate our π₯. Times four divided by four cancels out. And we found a second solution of π₯ equals negative 11 over four.

From here, in order to find out if one of these is an extraneous solution, weβll need to plug π₯ equals zero and π₯ equals negative 11 over four into our original equation. So letβs check π₯ plus nine, the square root of π₯ plus nine minus two π₯ equals three when π₯ equals zero. We have the square root of zero plus nine minus two times zero equals three. That would be the square root of nine minus zero equals three. So we say, does the square root of nine equal three? It does. π₯ equals zero is a valid solution.

Weβll need to do the same process for π₯ equals negative 11 over four. Negative eleven-fourths plus nine equals 6.25. Negative two times negative eleven-fourths equals five and a half. Bring down the five and a half. Two and a half plus five and a half equals eight. And eight does not equal three. This tells us that π₯ equals negative eleven-fourths is not a valid solution to our problem.

Because π₯ equals negative 11 over four is not a valid solution, we call it an extraneous solution.