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# IIT JEE hairy trig and algebra (part 1)

Video transcript

"Let ABC be a triangle,
such that angle ABC is equal to pi over 6." Let me draw it. So that's A, B, and
C. And they tell us that angle ACB, this angle
right over here, is pi over 6. I'm assuming it's
pi over 6 radians. Or we could view
that as 30 degrees. And then they tell
us, "Let a, b, and c denote the lengths of the sides
opposite to capital A, B, C, respectively." So this is a, this is b, and
this is c, right over there. The values of x, for which
a is equal to this thing, b is equal to this thing, and
c is equal to this thing is? So let's think about it. One, we need to find a
relationship between A, B, and C in this angle. And then once we have
that relationship, we can substitute them
with these expressions. And then essentially
try to solve for x. And the first thing you
want to think about is, OK, how can I relate
the length of b, this lowercase c, this lowercase
a with this angle here.? And what should pop
out into your brain is the law of cosines. Law of cosines. Law of cosines that
tells us that c squared is equal to a squared
plus b squared. It's essentially the Pythagorean
theorem with an adjustment right here, because this
isn't a right triangle. So a squared plus b
squared minus 2ab times the cosine of C. Now in this case,
angle C right here-- or I should say, in this case,
it's angle ACB-- this over here is going to be 30 degrees. So what's the cosine
of 30 degrees? And just in case you
forgot it, although I assume if you're going to
take the IIT exam, the JEE, you're going to need all the
help you need in terms of time. So this is 30 degrees. That is 1. Then this is 1/2, the side
opposite the 30-degree side. This will be 60 degrees. It's going to be a 30-60-90. And so this is square
root of 3 over 2. Cosine is adjacent
over hypotenuse, so square root of 3/2 over 1. So this, in this
example, it's going to be c squared is equal
to a squared plus b squared minus 2ab. And then this whole thing
is square root of 3/2. So times square root of 3/2. And these 2's cancel out. And so we get c squared is equal
to a squared plus b squared minus the square root of 3ab. Now the next thing I'd want
to do is substitute for a, I'd substitute with this
expression over here. For b, I want to substitute
with this expression over here. And then for c, I want to
substitute this expression right over here. But I want to be careful,
I want to make sure that it's in a form that I
can, in some way, simplify. Because if I just take this
a over here-- this x squared plus x plus 1-- if
I square it, I'm going to end up with
something to the fourth power. This b, I'm going to end up with
something to the fourth power. And then I have the square
root of 3 business over here. So let's see if we can at least
rearrange it, or at least maybe quarantine the square
root of 3, little bit, to make it at least
something that we might have some hopes of factoring,
or solving in some way. And I want to make a little
clarifying point here. In most Western exams,
if you're taking the SAT or you're taking any type of--
even at a math competition-- if you find yourself
grinding through math, multiplying huge polynomials
times each other, you're probably doing
something wrong. Most Western exams--
like the SAT-- you kind of have
to see the trick, you kind of have to see the
elegance in the problem. But once you see the
elegance, the problem is usually pretty
straightforward. In fact, if you're
grinding through math, it's usually a giveaway that
you're doing something wrong. On these JEE math
problems-- and I've only recently gotten
exposure to them. I obviously never took them,
not having grown up in India. But on these, you have
to kind of see the trick. And then you have to
grind through math, and you have to use every
trick at your disposal to make the grinding through
math a little bit easier. So the test writers for the
JEE are definitely a little, actually, a lot more
cruel than the test writers for most Western
exams that I've seen. So what I want to
do is, let's just isolate the square root of 3ab. So let's add it to both sides. So the square root
of 3ab plus-- so let's just add that--
square root of 3ab. And let's get the c squared
onto this side of the equation. So let me subtract c
squared from both sides of the equation. So minus c squared. And then minus c squared. And so I will have, on this
side, those cancel out. And I just have the
square root of 3ab is equal to a squared plus b
squared-- these cancel out-- minus c squared--
I'm just rearranging the whole thing--
minus c squared. All of that, well right
now, that's all it is. And now I can divide
both sides of this by ab. So if I divide
that by ab and then I divide the right-hand side
by ab, that cancels out. And I've kind of quarantined
that square root of 3. So I'll hopefully only
have integer stuff on this side of the equation. So I have the
square root of 3 is equal to a squared plus b
squared minus c squared. So let's figure out what
each of these things are. Let's expand them out,
actually, before we rewrite it. So let's figure out
what a squared is. So let's think about a squared. a squared is going
to be equal to-- and this is the part
where they clearly want to see whether you can
multiply multiple polynomials. That's going to be x squared
plus x plus 1 squared, or x squared plus x plus 1
times x squared plus x plus 1. And so this is going to be
1 times this whole thing, is x squared plus x plus 1. x times this whole
thing, is x plus x squared plus x to the third. And you may or might
not have noticed, I'm doing these a little faster
than I traditionally do them, just in the hope of
being able to finish this problem in a
reasonable amount of time. And then finally, x squared
times this whole thing, x squared times 1 is x squared. x squared times x
is x to the third. And then x squared times x
squared is x to the fourth. So a squared is going to
be x to the fourth plus 2x to the third plus 3x
squared plus 2x plus 1. So that's a squared. What's b squared? That looks a little bit
more straightforward. b squared, so that's b. So b squared is going to be--
well, that squared is just x to the fourth minus
2x squared plus 1. And then c squared. That's more straightforward. So at least they
weren't as painful as this. c squared is going to
4x squared plus 4x-- 2 times the product of these
two, so plus 4x-- plus 1. So the numerator of
our expression here. So let's rewrite this thing
that we were simplifying. It now becomes the square
root of 3 is equal to-- oh, and we don't want
to-- well, let's just do one step at a time. So it's equal to a squared,
which is this thing over here. So it is x to the
fourth-- actually, let me scroll down a little bit. It's this thing plus this thing,
minus this thing over here. So let me just do it. It's this thing
plus the blue thing, plus x to the fourth
minus 2x squared-- it'll be easier to add
them up this way-- plus 1. And then it's minus this
thing, minus c squared. So minus. So this is going to be a
minus 4x squared minus 4x. And then you have a minus 1. And then when you
add them all up-- so this will be the numerator
of this expression. We have the square root
of 3 is equal to-- so let me just do it down here. So we have 2x to the
fourth-- let's scroll to the left a little bit. So we have 2x to the fourth
plus 2x to the third. And then 3x squared minus 6x
squared is minus 3x squared. 2 minus 4 is minus 2x, and
then you have 2 minus 1 is-- then you have just plus 1. So the numerator, this thing
over here, simplifies to this. It's equal to 2x to the fourth
plus 2x to the third minus 3x squared minus 2x plus 1. And the denominator is ab. a is this thing over here. a is x squared plus x plus 1. And b, we saw it up
here, b is-- actually, we have it written up over
here. b is x squared minus 1. So b is x squared minus 1. Now at this point, we
need to solve for x. If we just multiply this
stuff times square root of 3 and went backwards, kind of
unsolved it back to the form that we had way up here, we'd
have a fourth degree polynomial with square roots of 3's in it. We still have a fourth
degree polynomial. But at least this one has
all integer coefficients, which at least
gives me some hope that I might be
able to factor it. I might be able to
find some roots. And if I can find
some roots that are common to the numerator
and the denominator, maybe I can reduce this
into a lower degree problem. So I'm going to leave you there. I'm pushing 10 minutes. In the next video, we're going
to try to factor this beast up here and see if it has any
common factors with this over here. And then try to solve for x.