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Solving square-root equations: no solution

Sal solves the equation √(3x-7)+√(2x-1)=0, only to find out that the single solution is extraneous, which means the equation has no solution.

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Video transcript

- [Voiceover] Let's say that we have the radical equation, the square root of three x minus seven plus the square root of two x 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 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 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 three x minus seven is going to be equal to this, the negative of the square root of two x 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. So we have to test our solutions at the end to make sure that they are actually valid for our original equation. But if you square both sides, on the left-hand side we're going to get three x minus seven, and on the right-hand side, a negative square's just a positive, and the square root of two x minus one squared is going to be two x minus one. Now let's see, we can subtract two x from both sides to get all of our xs on one side, so I'm trying to get rid of this. And we can add seven to both sides because I'm trying to get rid of the negative seven. So add seven to both sides. And we are going to get, we are going to get three x minus two x is x, is equal to negative one plus seven. x is equal to six. Now let's verify that this actually works. So, if we look at our original equation, the square root of three times six minus seven, minus seven, needs to plus, plus, the square root of two times six minus one, needs to be equals 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 eleven, 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, it's a 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 on the right-hand side and squared it, well it all boiled down to this, regardless of what starting point you started with, due to the exact same thing you would've gotten to that say, this 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. So this one, interestingly, has no solutions. And it'd actually 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 gets, that has a common intermediate step, but it's also fun to think about why, why this right over here is impossible.