IIT JEE hairy trig and algebra (part 3)

So I'm looking forward to the end of this problem as much as I'm guessing you are. And I think we're in the home stretch. We just have to simplify this to get the possible values for x. So let's simplify what we have here in the radical. So let's see. Square root of 3 minus 2. And we're going to square this whole thing. That's going to be square root of 3 squared, which is 3, minus 2 times their product, so that's minus 4 square roots of 3. and then plus 2 squared, so plus 4. And then we have minus 4 times-- square root of 3 times square root of 3 is 3. Square root of 3 times 1 is plus the square root of 3. And then negative 2 times square root of 3 is negative 2 square roots of 3. And then negative 2 times 1 is negative 2. So this is going to be equal 3 plus 4 is 7 minus 4 square roots of 3 minus-- so let's see, this 3 and this negative 2 over here, that becomes a 1. And then the square root of 3 minus 2 square roots of 3, that just becomes a minus square root of 3. So that just becomes negative square root of 3. My brain is starting to hurt. So you have a negative 4 times 1. So it's minus 4. And then a negative 4 times negative square roots of 3. So plus 4 square roots of 3. And thank God something has finally simplified here. And so that cancels out with that. So you have 7 minus 4 is equal to 3. So the whole thing under the radical sign is the square root of 3. So then this becomes our potential values for x. x is equal to 2 minus the square root of 3 plus or minus the square root of this thing, which simplified to the square root of 3, all of that over 2 square roots of 3 minus 4. So let's try to see what our answers could be. If we add a square root of 3-- so one answer is if we add the square root, if we go with the plus option, this would be 2 minus square roots of 3 plus square root of 3. So it would be 2. The square roots of 3 will just cancel out. 2 over 2 root 3 minus 4. We can divide the numerator and the denominator by 2. And we get 1 over the square root of 3 minus 2. And then to rationalize this, we want to multiply it times the square root of 3 plus 2 over the square root of 3 plus 2. We do that so that when we take the product here, it takes the form b squared minus a squared. So this is equal to the square root of 3 plus 2, all of that over-- square root of 3 squared is now 3-- 3 minus 4. So this is going to be equal to-- this part here is a negative 1. And so we could view it as negative, and I'll just write it as 2 plus the square root of 3. So that's one possible value of x. And the other possible value of x, if we subtract the square root of 3. So then on the numerator, we're going to have 2 minus-- square root of 3 minus the square root of 3, that's 2 minus 2 square roots of 3 over-- we could write this as negative 4 plus 2 square roots of 3. The first thing we might want to do is just divide the numerator and the denominator by 3. I swapped these down here to divide the numerator and the denominator by 2. So this is equal to 1 minus the square root of 3 over negative 2 plus the square root of 3. And now we can rationalize this by multiplying it by negative 2 minus the square root of 3 over negative 2 minus the square root of 3. And what do we get here? So this is going to be equal to-- in the denominator, we're going to have negative 2 times negative 2 is 4, 4 minus 3. So we just get a 1. So this is equal to 1 in the denominator. In the numerator, so let me just multiply 1 times negative 2 minus square root of 3. So it's negative 2 minus the square root of 3. And then the negative square root of 3 times negative 2 is plus 2 square roots of 3. And then negative square roots of 3 times negative square roots of 3 is plus 3. So you have 3 minus 2 is 1. And then 2 roots 3 minus the square root of 3 is plus the square root of 3. So this is kind of exciting. This was an unbelievably hairy problem. And if any of y'all know a faster way to do this problem, let me know. I've thought about it a lot. It doesn't seem like there is a faster way to do this problem. But my possible x-values are negative 2 plus the square root of 3 or 1 plus the square root of 3. Now, we have to be careful. If it wasn't enough that we got this far, we did so much work, we finally got these answers, we have to be very careful, because both of those are options. This is one of the options, and then this is the other option. And once again, these are one of those multiple correct choice answers, so you can actually select more than one answer here. Or you maybe need to select more than one answer. So you might, after doing all of that work, say oh, finally. OK, bam and bam. But be very careful. Remember the context of the problem. These sides had to be positive. And if you put this number in here, all the sides will not be positive. In particular, if x was 2 plus the square root negative of this value right here, this value over here is going to be, I don't know, 3 point something something. And if you took the negative of that, and you put it right here for C, you're going to get a negative distance for C. And C cannot be negative. It's the length of a side of a triangle. So A is not a valid choice. So these guys are really trying to trip people up. The correct answer is only B. That was tiring.