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# IIT JEE hairy trig and algebra (part 3)

Video transcript

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.