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IIT JEE hairy trig and algebra (part 2)

Video transcript
Welcome back. Where we left the last problem, we essentially found ourselves having to solve this equation right over here. And where we left off, we said, well, maybe, just maybe, we might be able to simplify this a little bit if we can find some common factors between the numerator and the denominator. And I want to do it here because if I multiplied this out, and multiply it times the square root of 3, and then subtract it or add it to that so I got just a straight-up polynomial, that polynomial is going to have the square root of 3 in it. It's going to be a really, really messy thing. And it's still going to be fourth degree. And at least this way, I have some chance of being able to solve it. At least I just have integer coefficients over here. And the only way that this is going to simplify is if it has any common factors with the stuff down here. And when you look over here, this right here is actually going to have complex roots if you did the negative b plus or minus the square root of b squared minus 4ac over 2a. The b squared minus 4ac over here is actually going to be negative. So it's not going to have any real roots. So that is not going to have any easy, real roots, but this is. The x squared plus 1, that has x plus 1 and x minus 1. Or another way to think about it is roots to this are negative 1 and positive 1. So if either of those are roots to this, then we're in business. And actually, just looking at this, there's a decent chance, because in general, if you want to think about the roots of really any n-th degree polynomial, the roots, if you think about them-- let me just write it over here. If I have a bunch of roots-- x minus r1, x minus r2, x minus r3, and then let's say we have a fourth degree, x minus r4. So let's say we have four roots. When you multiply this whole thing, you're going to have x to the fourth plus blah, blah, blah, blah, blah, all the way. And then the last term over here is going to be r1, r2, r3, r4. It's going to be the product of all of these. And so if you wanted to put this equation right or this expression in this form, you divide the whole expression by 2. Or you can kind of say, well, let's just look at this thing and divide it by 2. And then you would have x to the fourth plus 2x, all of that plus 1/2. And so this thing would have to, in this context, be equal to 1 over 2. And away, this is kind of a-- it's not a fail-proof technique, but it is one way to kind of discover roots, is you, say, OK, the roots could be the factors of this divided by the factors of that. And 1 is definitely a factor of both of those. And actually, if you think about it, negative 1 could be as well. So we feel good. But let's actually try it out. Instead of going through all of this argument, let's actually see if either 1 or negative 1 are factors of this thing up here. If they're not, then we just have to move ahead and multiply this thing out, multiply it by square root of 3, and then just solve. But we can hope. Usually the test writers aren't that big sadists. Otherwise, it would be impossible to solve. So I'm feeling good about this. So let's just try it with 1. So if we try 1 here, so let's put 1 here and see if we get 0 over here. So if we get 2 times 1 to the fourth plus-- so let me just write that. So 2 times 1 to the fourth is just 2, plus 2 times 1 to the third is just 2, minus 3 times 1 squared is just 3, minus 2 plus 1. So this is 4 minus 5 plus 1. That indeed equals 0. And let's try negative 1. So this is 2 times negative 1 to the fourth. That's still 2. Plus 2 times negative 1 to the third, so that's negative 2. Minus 3 times negative 1 squared, that's negative 3. And then minus 2 times negative 1, so that's plus 2 plus 1. So this is 0 minus 3 plus 2 plus 1. This also cancels out. This also gets us 0. So we now know that both x plus 1 and x minus 1 are factors of this thing up here. Or another way to think about it is their product is a factor of this thing up here. x squared minus 1 is a factor of this. So if we were to divide x squared minus 1 into this, we should get a reasonable answer. We shouldn't get a remainder, I guess I should say. So let's do that. So we're going to get a little practice doing polynomial long division. So if we divide x squared minus 1 into this thing up here, 2x to the fourth plus 2x to the third minus 3x squared minus 2x plus 1. So this is good practice on polynomial long division-- highest degree term going into highest degree term. x squared goes into 2x to the fourth 2x squared times. So let's put that in the 2x squared spot. 2x squared times x squared is 2x to the fourth. 2x squared times negative 1 is negative 2x squared. And now let's subtract this from up there. So we'll multiply by it negative 1 and then add it. And so this gives us 2x to the third. Negative 3 plus 2, so it's minus x squared minus 2x plus 1. And then x squared goes into 2x to the third 2x times. So this is plus 2x. 2x times x squared is 2x to the third. 2x times negative 1 is negative 2x. And then we can subtract this from there, or you could say add the negative. So let's just add the negative. It's easier to keep track of things. So those cancel out. You get negative x squared-- these cancel out-- plus 1. x squared goes into negative x squared minus-- or negative 1 times. Negative 1 times x squared is negative x squared. Negative 1 times negative 1 is positive 1. And indeed, we get no remainder. So we can write this thing up here. We can write it as the product of this. So it is equal to-- so this thing we can rewrite as 2x squared plus 2x minus 1 times x squared minus 1. So this times this is equal to this, so times x squared minus 1. So we just factored it. And so that simplifies things a good bit. It doesn't make it trivial, but it simplifies things a good bit. So now we can cancel that with that. And now at least we have things in second degree. We got rid of these third and fourth degree terms. So let's now rewrite things. So now let's multiply both sides of this equation by x squared plus x plus 1. So we're going to get the square root of 3-- so let me just write it out. So I'm going to get the square root of 3 times x squared plus the square root of 3 times x plus the square root of 3 times 1-- so I just multiplied the left-hand side by what we have left in the denominator-- is equal to what we have left in the numerator, because if we multiply the right side by this, it obviously cancels out. So this is going to be equal to 2x squared plus 2x minus 1. And now if we want to-- now let's see what we can do here. Well, we can subtract the stuff on the right side from the left-hand side. And so we would get the square root of 3 minus 2 times x squared plus the square root of 3 minus 2 times x-- just subtracting this stuff from that side-- plus the square root of 3 plus 1, right? Did I do that right? Square root of 3, I subtracted the x squared from that, so square root of 3 minus 2, square root of 3 minus 2. And you have the square root of 3. Subtracting a negative 1, I get a plus 1 over here. And since I subtracted all of this stuff from both sides of the equation-- this is just the left-hand side-- the right-hand side is now 0. I subtracted this from the right-hand side to get 0. So now we have-- this almost looks ridiculously simple. Even though I have these ridiculous radicals and all that, I have a straight-up quadratic equation right here, so I can use the quadratic formula. So the roots here, so x is going to be equal to negative b. This is b right here. So negative b, you can swap those. So it's 2 minus the square root of 3 plus or minus b squared. So let me just write it here. So the square root of 3 minus 2 squared minus 4 times a. So that's the square root of 3 minus 2 times c. Well, c is the entire constant term right over there. So it's times square root of 3 plus 1, all of that over 2a, so all of that over this business over here. So that's 2 root 3 minus 4. So we thought we were in the home stretch, but now we have to do some major radical simplification. So once again, these test writers are not pleasant people. So let's try our best to tackle this. So let's go into-- let's go inside the radical right over here. Let's go inside of the radical right over here. And you have the square root of 3 minus 2. Actually, I'm going to leave you there. I'm approaching 10 minutes, and frankly, I need a drink of water. This problem is maybe the most tiring problem I've ever done. I'll leave you there. We just have to simplify this in the next video.