 Welcome back to our lecture series, Math 1050, College Oddsworth for students at Southern Utah University. As usual, I'm your professor today, Dr. Andrew Misseldine. In this video, the start of lecture 18, we're going to continue our discussion of solving quadratic equations that we had done in lecture 17. Number 17 mostly focused on solving quadratic equations via factoring. But at the end, we saw that, the end of lecture 17, that is, we saw that using square roots can't be an effective tool for solving quadratic equations. But it seemed like there was a limit to when that's applicable or not. But we're going to see in this lecture that in fact, that limit is quite artificial, that we can solve any quadratic equation using the method of square roots. But it might require that we have to complete the square. In this video, I want to give us some examples of how one can complete the square. Consider the quadratic equation illustrated here on the screen, x squared minus 6x plus 8 equals 0. This is the so-called standard form of a quadratic equation. It looks like ax squared plus bx plus c equals 0, that is the right-hand side looks like 0. That's what we want in this type of situation. Now, the idea behind the complete square for a quadratic equation is to change this quadratic polynomial into one that's a perfect square trinomial. So you'll remember the factorization that we've seen previously, that if you take a plus or minus b squared, that's going to equal a squared plus or minus 2ab plus b squared. The advantage, of course, is that if we have a perfect square trinomial, it factors as a plus b squared or a minus b squared. And so if we have something like, say, x plus b squared equals c, then we'll get that x plus b equals plus or minus the square root of c and then x equals negative b plus or minus the square root of c. We can solve the quadratic equation if we can get a perfect square trinomial by the square root method. So how do we get that? How do we get there? Well, that's what completing the square is all about. So when one completes the square, we start off with what we have, right? We take a quadratic equation and what you're going to do is you're going to put just the square, just the x's on the left-hand side. So keep the x squared, keep the negative 6x, but this 8 is going to move to the other side. So we're going to have an x squared minus 6x is equal to a negative 8 if we move the 8 to the other side. Because when I say move 8 to the other side, I mean we're subtracting 8 from both sides. Now you might have noticed I left a big gap right here. That was not me doing something weird. That actually is the fact that when we complete the square, there's a third term we're looking for, right? So we have the a squared, we have the 2ab, but we don't have the b squared. That's what we're trying to look for. So imagine we have this large party going on and we have sitting at the head at the table our guest of honor, but our guest hasn't arrived yet. Well, do we just let anyone sit in the guest of honor seat before she arrives? Of course not. We leave that seat open so that when she gets here, we can place her in this location. But to complete the square, we have to find this guest of honor, but we have to identify who it is. So looking at this perfect square trinomial formula, we have our perfect square right here. We need a second perfect square, right? So this right here, this a squared is typically gonna be an x, that's great. So that means the middle term, this 2ab should look like 2x times something. So we have to identify what that something is. So looking at the middle number, and it doesn't matter whether it's plus or minus, that's not too consequential. We have to look at this middle number and we're gonna take half of that. So looking at the six right there, to take half of it just means to multiply it by one half. So we take half of six, which is equal to three. This is gonna give us that b value, but the guest of honor is gonna be b squared. So we have to then square this number, b squared is gonna equal three squared, which is equal to nine. And that is who we're gonna add in as our guest of honor. We need to add in the nine. But as this is an equation, what's good for the goose is good for the gander. We have to make sure we add nine to both sides of the equation so that it remains balanced. So this will then of course give us x squared minus six x plus nine on the left-hand side. And then on the right-hand side, you get negative eight plus nine, which is excuse me, equal to one, like so. The left-hand side you'll then recognize x squared minus six x plus nine. The whole point of adding the nine was that now the left-hand side is a perfect square trinomial. It'll factor as x minus three squared. So you end up with this x minus three squared. This is equal to one. Now notice where did this x minus three squared come from? Well, it's gonna be a perfect square because that's what we were trying to do. We're trying to complete the square here. You get an x by taking the square root of the x squared. You're gonna get a three by taking the square of nine. Or in other words, this was the b value that we found earlier. And then this is gonna be a negative three because we have a negative six before. Basically, you take the coefficient in the middle. That's going to give you, you take half of it. That's the number that appears here. Half of negative six is negative three. Now the reason why we needed the nine wasn't necessarily to get the factorization so that we knew what had to be added to both sides of the equations because the plus nine on the left-hand side kind of disappears when we factor it. But the plus nine on the right-hand side did have a consequence. Negative eight plus nine gives us one. Once we have the square completed and factored on the left-hand side, we can then solve this equation by square roots. We get x minus three equals plus or minus the square to one. Square to one, of course, is one itself, so you get plus or minus one. And thus, x equals three plus or minus one, which is the same thing as four. I should say that three plus one is four and three minus one is then a two, which is the solution to this equation here. Now I want to confess that if we had looked at the original equation, we could have actually factored this, or we could have solved this by factoring, right? Factors of eight that add up to be negative six, you could take x minus four and x minus two equal to zero, which would then give you x equals four and two like we saw just right here. So in terms of factoring, well, in regard to factoring, completing the square is not necessarily meant to be a superior faster method. I should mention, of course, that we'll see in forthcoming examples that factoring does have its limitations, but in situations like this, x squared minus six x plus eight, I would solve this one by factoring. This is just an example to illustrate how one completes the square. Let's look at some other examples. Now again, for the sake of practice, we're going to complete the square here. Let's move the four to the other side. So we get x squared plus five x is equal to negative four. Remember to leave a seat open for our guest of honor. We're gonna take half of the middle coefficient, half of the five. When we take half of it, the fact that it's an odd number doesn't make much of a difference. Half of five is gonna be five halves. That's our b value. We have to square it and we get 25 halves, sorry, 25 fourths, square the top, square the bottom. So we add 25 fourths to the left. We have to also do that to the right, like so. Now the left-hand side will factor since we've completed the square, it'll factor is x, because we have an x squared right here. Then the next number is gonna be our b value, half of the middle coefficient, keeping the same sign. So we get plus five halves squared. And then on the right-hand side, since we have negative four plus 25 fourths, we should probably write the negative four as negative 16 fourths plus 25 fourths, which that's gonna give us a nine fourths, like so. Then taking the square root of both sides, we get x plus five halves is equal to plus or minus the square root of nine fourths. Remember that you take plus or minus. Now in this situation again, you take the square root of nine fourths, that's a perfect square, that'll just become three halves. And so x will be equal negative five halves if you move that to the other side, plus or minus three halves. Now if we take plus three halves, that's gonna be negative two halves. If we take minus three halves, it's gonna be negative eight halves, thus giving us the solution negative one and negative four. Which of course, if we compare that to how we might have done this by factoring, if you try to factor this thing instead, factors of four that have to be five, you get x plus four and then x plus one, which then gives you the same solutions, negative one and negative four. So again, factoring of both these situations turned out to be much more advantageous. But this is something you're gonna see in general, that if you solve a quadratic equation by completing the square, if you ended up with like a perfect square right at this moment we took the square root, it turned out you didn't have to complete the square. You could have actually done it by factoring, but hindsight's always 20-20, right? Let me show you one more example of completing the square in this video here. This time, consider the equation four x squared minus 32x plus 23. Following the same strategy as before, we're gonna move the 23 to the other side of the equation. So we get four x squared minus 32x is equal to negative 23. Now in this situation, whenever you're leading coefficient, that is the coefficient of x squared is anything other than one. I want you to factor it away from the x square but also factor it away from the x. And so if you factor out the four, you're gonna get four times x squared, that's easy enough. Factoring the four away from negative 32, of course, just means that you're taking negative 32 divided by four. And so you could negative 8x, leave your space open for your guest of honor, negative 23. And then at this moment, we can either divide by four on both sides or we can leave it there for a moment. We'll have to do it eventually. But for the sake of this example, I'm gonna leave it where it is. Let's identify who the guest of honor is gonna be. So we need to take half of negative eight, which if we take half of negative eight, that's gonna give us negative four, who's our b value? And thus, b squared is gonna equal 16. Negative four times negative four is 16. So we're gonna add 16 inside because that's our guest of honor. Then we have to add 16 to the other side but just a second, I want you to be aware that when you actually distribute, if you distribute this four back through, you're gonna get four times 16, which means you actually added four times 16 on the left-hand side. So to correct that, you have to add 16 times four on the right-hand side. You have to add your guest of honor times whatever the coefficient of the leading term was. You had to factor that out. If you had divided by four earlier, that step can kind of be skipped because we get a fraction right here. But on the other hand, you'll have to get it done eventually when you start finding a common denominator for these things. So this is why I kind of didn't bother with the division. I'm trying to avoid fractions until it's absolutely necessary. So then carrying forward with this, we have four times X minus four quantity squared because the left-hand side is now a perfect square, X squared minus eight X plus 16. Now the right-hand side, we're gonna have negative 23 plus 16 times four. Well, multiplying that out, we should get, what is that gonna be, 52? No, that's not right, what am I talking about? Oh boy, 16 times four, you're gonna get 40 plus 24, so that's 64. And then subtract 23 from that, you should end up with 41. And so now we wanna solve this equation via the method of square roots. Now I would divide both sides by four. Fractions can't be avoided forever. So divide both sides by four, we get X minus four squared is equal to 41 fourths. Take the square root of both sides, we end up with X minus four is equal to the square root of 41 over four plus or minus. Now, while the denominator is a perfect square, the square root of four of four is two, the square root of 41, you're not gonna be able to reduce that at all. 41 is not a perfect square, and in fact the largest divisor of 41 that is a perfect square is one. So no type of simplification of the square root is even possible here. Adding four to both sides, we see that the final solution would be four plus or minus the square root of 41 over two. Now if you prefer this to have to be a common denominator, you're gonna get eight plus or minus the square root of 41 over two. And this is two different solutions, of course, there's eight plus root 41 over two and eight minus root 41 over two. But however you wanna slice it, these two solutions are actually gonna be irrational numbers. The square root cannot be simplified out by taking perfect squares. So you actually get two irrational solutions on this example right here. Now there's some consequences of that. I mean, they are still real numbers, of course, but their decimal expansions will be non-repeating. And I think, for the most part, you're gonna write your answers as exact when you have square root of 41. Don't try to approximate unless you are instructed to round it to three decimal places or whatever. If there's no instructions to round, then you should always write your answers as exact in this situation. But the other thing I wanna mention here is that because the answers turned out to be irrational numbers, that actually indicates that we could not have factored this. So the completing the square was sort of necessary in this situation, factoring wouldn't have worked. If you look at factors of four times 23, you look through all the options, you're not gonna find any magic pair that adds up to be negative 32. And so the advantage of completing the square here is that it could solve the problem when factoring could not. And also completing the square is completely algorithmic. There's no point in the process where a decision has to be made. There's no search for magic pairs or anything. You can just kinda chug it through the machine, turn the crank, and that'll get you the solution each and every time. And that does kinda provide a nice advantage to completing the square over factoring. I like factoring because it can be very quick, but it also doesn't work all the time. Completing the square will always work assuming its operator knows what they're doing. And that's why we like practicing completing the square.