Intro to square-root equations & extraneous solutions

In this video, we're going to start having some experience solving radical equations or equations that involve square roots or maybe even higher-power roots, but we're going to also try to understand an interesting phenomena that occurs when we do these equations. Let me show you what I'm talking about. Let's say I have the equation the square root of x is equal to 2 times x minus 6. Now, one of the things you're going to see whenever we do these radical equations is we want to isolate at least one of the radicals. There's only one of them in this equation. And when you isolate one of the radicals on one side of the equation, this one starts off like that, I have the square root of x isolated on the left-hand side, then we square both sides of the equation. So let us square both sides of the equation. So I'll just rewrite it again. We'll do this one slowly. I'm going to square that and that's going to be equal to 2x minus 6 squared. And squaring seems like a valid operation. If that is equal to that, then that squared should also be equal to that squared. So we just keep on going. So when you take the square root of x and you square it, that'll just be x. And we get x is equal to-- this squared is going to be 2x squared, which is 4x squared. It's 2x squared, the whole thing. 4x squared, and then you multiply these two, which is negative 12x. And then it's going to be twice that, so minus 24x. And then negative 6 squared is plus 36. If you found going from this to this difficult, you might want to review multiplying polynomial expressions or multiplying binomials, or actually, the special case where we square binomials. But the general view, it's this squared, which is that. And then you have minus 2 times the product of these 2. The product of those two is minus 12x or negative 12x. 2 times that is negative 24x, and then that squared. So this is what our equation has I guess we could say simplified to, and let's see what happens if we subtract x from both sides of this equation. So if you subtract x from both sides of this equation, the left-hand side becomes zero and the right-hand side becomes 4x squared minus 25x plus 36. So this radical equation his simplified to just a standard quadratic equation. And just for simplicity, not having to worry how to factor it and grouping and all of that, let's just use the quadratic formula. So the quadratic formula tells us that our solutions to this, x can be negative b. Negative 25. The negative of negative 25 is positive 25 plus or minus the square root of 25 squared. 25 squared is 625, minus 4 times a, which is 4, times c, which is 36, all of that over 2 times 4, all of that over 8. So let's get our calculator out to figure out what this is over here. So let's say so we have 625 minus-- let's see., this is going to be 16 times 36. 16 times 36 is equal to 49. That's nice. It's a nice perfect square. We know what the square root of 49 is. It's 7. So let me go back to the problem. So this in here simplified to 49. So x is equal to 25 plus or minus the square root of 49, which is 7, all of that over 8. So our two solutions here, if we add 7, we get x is equal to 25 plus 7 is 32, 32/8, which is equal to 4. And then our other solution, let me do that in a different color. x is equal to 25 minus 7, which is 18/8. 8 goes into 18 two times, remainder 2, so this is equal to 2 and 2/8 or 2 and 1/4 or 2.25, just like that. Now, I'm going to show you an interesting phenomena that occurs. And maybe you might want to pause it after I show you this conundrum, although I'm going to tell you why this conundrum pops up. Let's try out to see if our solutions actually work. So let's try x is equal to 4. If x is equal to 4 works, we get the principal root of 4 should be equal to 2 times 4 minus 6. The principal root of 4 is positive 2. Positive 2 should be equal to 2 times 4, which is 8 minus 6, which it is. This is true. So 4 works. Now, let's try to do the same with 2.25. According to this, we should be able to take the square root, the principal root of 2.2-- let me make my radical a little bit bigger. The principal root of 2.25 should be equal to 2 times 2.25 minus 6. Now, you may or may not be able to do this in your head. You might know that the square root of 225 is 15. And then from that, you might be able to figure out the square root of 2.25 is 1.5. Let me just use the calculator to verify that for you. So 2.25, take the square root. It's 1.5. The principal root is 1.5. Another square root is negative 1.5. So it's 1.5. And then, according to this, this should be equal to 2 times 2.25 is 4.5 minus 6. Now, is this true? This is telling us that 1.5 is equal to negative 1.5. This is not true. 2.5 did not work for this radical equation. We call this an extraneous solution. So 2.25 is an extraneous solution. Now, here's the conundrum: Why did we get 2.25 as an answer? It looks like we did very valid things the whole way down, and we got a quadratic, and we got 2.25. And there's a hint here. When we substitute 2.25, we get 1.5 is equal to negative 1.5. So there's something here, something we did gave us this solution that doesn't quite apply over here. And I'll give you another hint. Let's try it at this step. If you look at this step, you're going to see that both solutions actually work. So you could try it out if you like. Actually, try it out on your own time. Put in 2.25 for x here. You're going to see that it works. Put in 4 for x here and you see that they both work here. So they're both valid solutions to that. So something happened when we squared that made the equation a little bit different. There's something slightly different about this equation than that equation. And the answer is there's two ways you could think about it. To go back from this equation to that equation, we take the square root. But to be more particular about it, we are taking the principal root of both sides. Now, you could take the negative square root as well. Notice, this is only taking the principal square root. Going from this right here-- let me be very clear. This statement, we already established that both of these solutions, both the valid solution and the extraneous solution to this radical equation, satisfy this right here. Only the valid one satisfies the original problem. So let me write the equation that both of them satisfy. Because this is really an interesting conundrum. And I think it gives you a little bit of a nuance and kind of tells you what's happening when we take principal roots of things. And why when you square both sides, you are, to some degree, you could either think of it as losing or gaining some information. Now, this could be written as x is equal to 2x minus 6 squared. This is one valid interpretation of this equation right here. But there's a completely other legitimate interpretation of this equation. This could also be x is equal to negative 1 times 2x minus 6 squared. And why are these equal interpretations? Because when you square the negative 1, the negative 1 will disappear. These are equivalent statements. And another way of writing this one, another way of writing this right here, is that x is equal to-- you multiply negative 1 times that. You get negative 2x plus 6 or 6 minus 2x squared. This and this are two ways of writing that. Now, when we took our square root or when we-- I guess there's two ways you can think about it. When we squared it, we're assuming that this was the only interpretation, but this was the other one. So we found two solutions to this, but only 4 satisfies this interpretation right here. I hope you get what I'm saying because we're kind of only taking-- you can imagine the positive square root. We're not considering the negative square root of this, because when you take the square root of both sides to get here, we're only taking the principal root. Another way to view it-- let me rewrite the original equation. We had the square root of x is equal to 2x minus 6. Now, we said 4 is a solution. 2.25 isn't a solution. 2.25 would've been a solution if we said both of the square roots of x is equal to 2x minus 6. Now you try it out and 2.25 will have a valid solution here. If you take the negative square root of 2.25, that is equal to 2 times 2.25, so that is equal to 4.5 minus 6, which is negative 1.5. That is true. The positive version is where you get x is equal to 4. So that's why we got two solutions. And if you square this-- maybe this is an easier way to remember it. If you square this, you actually get this equation that both solutions are valid. Now, you might have found that a little bit confusing and all of that. My intention is not to confuse you. The simple thing to think about when you are solving radical equations is, look, isolate radicals, square, keep on solving. You might get more than one answer. Plug your answers back in. Answers that don't work, they're extraneous solutions. But most of my explanation in this video is really why does that extraneous solution pop up? And hopefully, I gave you some intuition that our equation is the square root of x. The extraneous solution would be valid if we took the plus or minus square root of x, not just the principal root.