when a quadratic equation has an infinite root.

today I'd like to look at an interesting thing that happens when you use the quadratic formula to solve a linear equation well we can take any linear equation BX + Cal 0 and then just add a 0 x^2 to it and now we have something that looks like a quadratic equation of course it's not exactly a quadratic equation but what if we were to think of it as a quadratic equation and then try to use the quadratic formula to solve this what would we end up with well the answer is kind of surprising and leads us to look at a nice area of math that perhaps you haven't seen before okay so how can we approach this and I guess before we start approaching this I'd like to maybe simplify the setup a little bit by saying that let's take B and C both to be nonzero here so I can can think about this equation here as the limit as a approaches zero of the equation ax^2 + BX + C = 0 so of course talking about the limit of an equation doesn't really make a ton of sense so perhaps what we should really be talking about is the limit of the solutions to this equation and of course if we're taking the limit as a approaches zero that means a is never zero because limits are all about deleted neighborhoods if you will so let's go ahead and solve this we know that the solution to this is X = B +us < TK of B ^2 - 4 a c all over 2 * a and what we'd like to do now is well take the limit of this expression right here this solution with a approaching zero but notice we've got a plus minus here I think it's not too hard to do this all at once but we're going to separate it out into two cases so let's look at the first case when we look at the plus so we've got the limit as a approaches Zer of B plus the square < TK of B 2 - 4 a c all over 2 a so how might we take this limit well observe that it's an indeterminate form of type 0 over Z notice the numerator cancels to zero because we've got netive B plus well the square root of B which is plus b over 2 a so perhaps well what should we do here well I think maybe multiplying by the radical conjugate is not a terrible idea so let's do that so the radical conjugate will be B plus the square < TK of b^2 - 4 a c over B+ the sare < TK of b^2 - 4 a c because we have to multiply by the radical conjugate in the numerator as well as the denominator I guess maybe the traditional radical conjugate would change this plus to a minus but if we just change this minus to a plus it's it's like doing the same thing and multiplying by negative 1 and I think it's nicer okay so this is going to give us the limit as a approaches zero of so multiplying this out will give us b^2 + b^2 - 4 a c over 2 a * b + the < TK B ^2 - 4 a c but now let's observe that some things cancel in the numerator this b^2 cancel with these b^2 and now our A's will cancel so this a will cancel with this a and then finally this two will cancel this four down to a two and then we have a minus sign that we can pull out as well so let's see what we have oh and we might as well take a approaching zero here because the only a left doesn't give us a zero in the denominator or anything so it's okay so let's see in the numerator we're left with - 2 * C and in the denominator we're left with B plus the square < TK of b^ s but the square root of b^ squ isn't B it is the absolute value of B so we have B plus the absolute value of B so if we're keeping track of solutions over here our first solution to this equation right here is 2 C over B plus the absolute value of B okay so now let's look at our second solution that we gain from taking our quadratic formula with a minus sign so the same thing is going to go here this limit is going to be very very similar so we'll have B minus < TK of b^2 - 4 a c all over 2 a we're going to do the same thing by multiplying by the radical conjugate get so now let's maybe take b + < TK b^2 - 4 a c all over the same thing because we've got to multiply in the numerator and the denominator we don't want to change anything okay so let's see where that leaves us so we'll have the limit as a is approaching zero we have a b^ 2- b^ 2 uh + 4 A a c after Distributing that minus sign through and this is going to be all over 2 * a time we've got this B plus the < TK of b^ 2 - 4 * a * C okay so that's looking good I think but now we can do some more cancellations just like we did before this b^2 will cancel this b^ s and then this a will cancel this a and then also the two two we'll cancel the four down to a two and now we're left with 2 C over so we'll have Nega B plus the square < TK of b^2 oh but that's going to be the same thing as we had before well we got to keep in mind of course that the square root of B squ is the absolute value of B so we have negative B plus the absolute value of B now I'm going to bring that minus sign up just to make it look more like this and now we'll have this is our second solution -2 C over B minus the absolute value of B but observe that this denominator is going to be zero all of the time well at least one of these denominators is going to be zero all of the time and that's because if you take this first one and we can simplify it two different ways if B is bigger than zero well then that means absolute value of B is the same thing as B those add up to 2 B which gives us the solution of minus C over B but if B is less than zero then B plus absolute value of B is zero and we're left with a zero in the denominator and I'm going to think about this as Infinity because we've got non zero in the numerator and zero in the denominator of course maybe we're not being as careful as we could be here but I'm going to think about that as a solution at infinity and then well we can do the same kind of game over here so observe over here if B is bigger than zero well then this stuff cancels and we get our solution at Infinity whereas if B is less than zero then that's going to give us our C over B so regardless of what happens notice that these are two disjoint possibilities we get two solutions this solution of netive c overb and then this solution at Infinity so let's see if we can make sense of this using another branch of math called projective geometry so in order to do that we need to consider the following object this is called the real projective plane and I'm going to write it as rp2 and what it is it's the quotient of three space real three space well almost all of real three space everything in real three space except for the origin modulo an equivalence relation so that means it's made up of equivalence classes and how do we Define the equivalence relation well it's like this so x y z is equivalent to rst if and only if there exists a Lambda which is non zero that makes X = to Lambda r y = to Lambda s and Z = to Lambda T so in other words one point is like a multiple of another point so now let's observe the following so if Z is non zero oh and I should point out before we really dive into this more here we'll use the following notation for elements inside of rp2 so in other words the equivalence classes so it'll be this ordered triple but instead of colons or sorry instead of commas we'll use colons okay so if Z is non zero observe that the point x colon y colon Z is the same thing as X over z y over Z1 because we can just use our scaling factor of Lambda as Z if Z is non zero well then we can just divide by Z and we get an equivalent point so that motivates us to have the following de composition of the real projective plane so we can take rp2 and decompose it into well this first part which is everything of the form X y1 where X and Y are allowed to be anything at all that's because well that's not going to include the origin uh which is the only thing that we're not allowing because this third coordinate is already not zero but there will be a case when the third coordinate is is zero and that's going to be everything else so I'm going to write that like this so X colon Y colon 0 and here X and Y are both real but they're also not both zero so let's maybe write it like this so not both zero and then well let's observe that this first bit is really just a copy of R2 because it has all of the degrees of freedom of R2 now this second bit well that's not really a copy of R2 because it doesn't include the origin and in fact if it doesn't include the origin then what we could really do here is well when Y is non zero we could divide by Y and we'll get uh you know X colon one colon zero and then the same thing for x so this is actually a copy of the real projective line and we actually think about this as the line at Infinity um but it's not like a normal line it's a real projective line which has an infinite Point itself we're not going to go into that super carefully okay so now let's look at solutions to this in this world over here so now it may not seem like anything interesting is going to happen but something will happen that will link us um to what we saw before in our you know limiting procedure so I'm going to drop the 0x2 because we won't need to worry about it in this case so solutions to BX + C = 0 in R P2 so first off one thing that we need to do is homogenize this equation so this is is a linear equation but it's not a homogeneous linear equation because notice we've got a linear term the BX and then we've got a degree zero term the C and we'll homogenize this by replacing x with X over Z that's going to give us B * X over Z + C = 0 but of course we can multiply through and get BX + c z equals 0 but now that has infinitely many solutions we could take X to be equal to Lambda * C and then we could take Z to be equal to let's see what it would be it would be minus Lambda * B and observe that here Y is really allowed to be anything we want because it's not represented inside of the equation okay so now let's take this and write it as a ordered triple in our real projective plane so that means all of our Solutions are the following form Lambda C colon Y where Y is like some number and then colon minus Lambda B okay nice but now this like kind of naturally breaks apart into two pieces maybe the first piece that this breaks apart into is the case when Lambda is equal to zero so well if Lambda is equal to Z then observe that that means that Y is not allowed to be zero but if Y is not allowed to be Z then we can divide by Y and scale this down to the point 0 colon 1 colon 0 but notice that's one of the points at infinity or that's on the line at Infinity so I'll just put here in quotes that this is infinity so this is like building another way this infinite solution that we built via the quadratic formula now of course if Lambda is not equal to zero well then we can divide this whole thing by minus Lambda B and we'll get the following so let's maybe do that right here so like I said this is when Lambda is not zero so that will give us something like this so we'll have Min - c over B colon Y and then colon one and now that's going to be well notice that's also infinitely many solutions but it's infinitely many solutions along maybe one of the axes in our original copy of R2 so that would maybe represent in a loose sense this solution over here which is minus C over B so there you have it two ways of looking at how we can get two solutions to a linear equation AKA how to use the quadratic formula to solve this equation where the coefficient of x s is zero and that's a good place to stop

Channel: Michael Penn

Views: 91,006

Keywords: math, mathematics, number theory, abstract algebra, calculus, differential equations, Randolph College, randolph, Michael Penn

Id: wTEYApUX-N8

Length: 16min 46sec (1006 seconds)

Published: Mon May 06 2024