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IIT JEE algebraic manipulation

Video transcript
What I want to do in this video is something a little unconventional. Because I tried this problem before I started this video, and I kept getting an answer that isn't one of these answers over here. So assuming that the exam writers did this properly-- and this is a serious exam. It's to get into a very prestigious set of universities in India, so I'm assuming they didn't make any major mistakes here. I want you to tell me where I'm making the mistake. So let's just try to do the problem. Let p and q be real numbers such that p does not equal 0, p to the third does not equal q, and p to the third also does not equal negative q. If alpha and beta are nonzero complex numbers satisfying alpha plus beta is equal to negative p and alpha cubed plus beta cubed is equal to q, then a quadratic equation having alpha/beta and beta/alpha as its roots is. Did I just call it b up here? It's beta. So what could that quadratic be? Well, these are the two roots, so that's pretty straightforward. We could say that the quadratic, a possible quadratic here, would be x minus alpha/beta times x minus beta/alpha is equal to 0. Clearly, the roots of this equation right here are alpha/beta and beta/alpha. Either one of those would satisfy this equation. Now let's multiply this out. x times x is x squared. x times negative beta/alpha is negative beta/alpha x. Then you have negative alpha/beta x, so negative alpha/beta x. And then you have negative alpha/beta times negative beta/alpha. The alpha's cancel out. The beta's cancel out. The negatives cancel out. And you get plus 1 is equal to 0. And then let's see, we might be able to simplify this a little bit. If we multiplied everything times alpha beta, what will happen? So I'm just going to multiply both sides of this equation times alpha beta. I don't like having things in denominators. Clearly, the right-hand side is still going to be 0. The left-hand side now becomes alpha beta x squared. Alpha beta times beta/alpha, the alphas cancel out. You have a beta times a beta. So it's negative beta squared x minus-- the betas cancel out. And now you have an alpha times an alpha minus alpha squared x. And now you have plus alpha beta is equal to 0. And then we can combine these terms over here. So we have alpha beta x squared minus-- you have an alpha squared plus a beta squared x-- all right, I just took the negatives out of it-- plus alpha beta is equal to 0. All right. So everything I did is just kind of-- I just said, OK, these are the roots. Let me just multiply it out, and I got something in terms of alpha and beta. Now when you look at it, their choices, clearly they want answers in terms of p's and q's. So let's see if we can convert this into p's and q's. So a good place to start, they tell us that alpha plus beta is equal to negative p, or negative p is equal to alpha plus beta. And so if we squared that, what I want to do is somehow relate it to q as well. And so maybe if we eventually take it to the third power, we can get some of these terms over here and relate it to q. And also, when we square it, I think we'll be able to relate some of these things to it as well. So let's square both sides of this equation. So you have negative p squared, which, because the negative canceled out, this is the same thing as negative 1 times p squared, which is negative 1 squared times p squared, which is just p squared, is equal to this expression over here squared, which is just alpha squared plus 2 alpha beta plus beta squared. Fair enough. And already we have something interesting showing up over here. We can express alpha squared plus beta squared in terms of p squared and alpha beta. And then we simplify things a little bit. Well, we'll just keep going. Let's just take this to the third power. So if we have negative p to the third power, which is the same thing as the negative of p to the third, if you multiply this thing times negative p, you're going to get the negative sign p to the third. And this is equal to-- we could just multiply again, do the distributive property. Or if you kind of remember Pascal's triangles, you remember the coefficients on it, or the binomial theorem, this would be a faster way to do it. I'll do a Pascal's triangle really quick here. So you have 1. 1, 1 for binomial. 1, 2, 1, when you take the square of a binomial, 1, 2, 1. And then when you take that to the third, when you go to the third power, it's 1, 3, 3, 1. Bunch of videos on this, especially in the videos on the binomial theorem. So this to the third power, this thing right here to the third power, is going to be alpha to the third plus 3 alpha squared beta plus 3 alpha beta squared plus beta to the third. I just decreased the exponent on the alpha each time and started increasing the exponent on the beta each time. Or you could just multiply this times this, and you should also get this. And now this is interesting because this is expressing everything in terms of-- let's see if we can simplify this a little bit. This is equal to-- so let me write this-- we have negative p to the third, which is equal to negative p to the third, just a difference of the parentheses, is equal to-- we have this alpha cubed plus the beta cubed. That's those two terms right over there. And I separate those out because those are our q. Those are our q. And then I have-- let's see, over here I can factor out a 3 alpha beta, so plus 3 alpha beta. And what do I get? If I factor out a 3 alpha beta here, I'm just left with an alpha on this term. And if I factor out a 3 alpha beta here, I'm just left with a beta. I am just left with a beta. So what does this simplify to? So this is going to simplify to-- this right here is equal to our q. So this is equal to q plus 3 alpha beta. And what's alpha plus beta? Well, they already told us what alpha plus beta is. Alpha plus beta is negative p. It is negative p. So let's see what we can do over here. And this right here is the same thing as negative p to the third. So here we can solve for alpha beta, and so we'll have alpha beta in terms of p's and q's. So that'll be good. So we can substitute the alpha beta's with whatever we get in terms of p's and q's. And then this thing we can express in terms of a p and an alpha beta, which means we can express it in terms of p's and q's. So let me do what I'm talking about. Let me do it I'm talking about. So let's just solve, first of all, for alpha beta. So if we solve for alpha beta, first we want to subtract q from both sides of this equation. So you have negative p to the third minus q is equal to-- we have this negative sign over here, so it's equal to negative 3 alpha beta p. And I want to solve for alpha beta, so let me divide both sides by negative 3. And so we get, on the right-hand side, we get alpha beta is equal to, if you divide-- so let me just write it out because I don't want to make a careless mistake. So you have negative p to the third minus q over negative 3p. Now if we multiply the numerator and denominator by a negative sign, we get p to the third plus q over 3p is equal to alpha beta. So we can now express this and this term in terms of p's and q's. Now what about this one right here? So what is alpha squared plus beta squared? And we see right up here, alpha squared, that's alpha squared plus beta squared. So if we subtract 2 alpha beta from both sides of this equation, we get alpha squared plus beta-- let me do this in a different color-- we get alpha squared plus beta squared is going to be equal to p squared minus 2 alpha beta. All right? I just subtracted 2 alpha beta from both sides. And so this is going to be equal to p squared minus 2 times this thing, minus 2 times p to the third plus q over 3p. And what is this equal to? This is equal to-- let's put on a common denominator-- p squared over-- this is the same thing as 3-- my cell phone rang. I had to pause it, sorry. So this is equal to 3p to the squared-- if I want to put it over 3p-- is equal to 3p to the third over 3p. Let me make it clear. This term right here is this term over here if you were to simplify it. The 3's cancel out. You just have p squared. And then you have minus-- I'm just going to distribute the negative 2-- minus 2p to the third minus 2q, all of that over 3p. So it is equal to 3p cubed minus 2p cubed is equal to p cubed minus 2q, all of that over 3p. So now we can write this entire expression in terms of p's and q's. This first term over here, alpha beta, this is p cubed plus q over 3p, and we want to multiply that times x squared. And then you have minus alpha squared plus beta squared. We just figured out that alpha squared plus beta squared is this thing over here. So minus-- I'll do it in a different color, I'll do it in green-- minus this thing, p to the third minus 2q over 3p. And that's going to be multiplied by an x, and then plus alpha beta again. So alpha beta is p to the third plus q over 3p. Now, when we look at any of the choices up here, it looks simplified. We don't see any 3p's in the denominator. So let's see if we can simplify that. Oh, and we don't want to lose-- we had an equals 0 here the whole time. So this is equals 0. So when you multiply both sides of this equation by 3p, what do you get? This first term becomes p to the third plus q times x squared. The second term becomes minus p to the third minus 2q x. And then this last term becomes plus p to the third plus q, and this is going to be equal to 0. Now, let's see how this compares to the choices. And I think I actually did not make my careless mistake in this video. I actually got the right answer. Let me see. It shows you, when you do it carefully, you're less likely to make errors. OK. So let's see. How does this match up to what I have over here? Well, this looks exactly. I'm very excited because I tried this twice before actually making the video, and I kept making a careless mistake. But this time I didn't make the careless mistake, and I got B, p to the third plus q x squared minus p to the third minus 2q x plus p to the third plus q. So that right there, that right there is our answer. So I take back what I said at the beginning of video. I didn't need your help. But I guess just the process of explaining it was the help that I needed. Anyway, hopefully you found that useful.