 This video will talk about solving quadratic equations and inequalities. There are four ways that we know how to do these, and the first way is to do it graphically. We can find the zeros, which are the solutions to the quadratic set equal to zero, which are actually just the x-intercepts when we look on the graph. They want us to find it graphically, but let's talk about factoring real quick. So if I had this problem, I would want to rewrite it as x squared minus 3x minus 18 equal to zero. And then I would take x and x and factors of 18 that would add up to three or subtract a three since there are going to be opposite signs here, this negative 18. It would be a negative six and a positive three, and then you could set those equal to zero, x minus six equals zero, so x equals positive six, and x plus three equals zero, so x equals negative three. Now let's see if we get the same values, it's a negative three. If we graph, if we want to find the x-intercepts or the zeros, we need to use this equation right here in our calculator. We're going to put in x squared minus 3x minus 18, and then we want a standard window because this is a simple graph. We're hoping it's going to be between six and negative three, which works. We need to do second trace, and we just want to find the zeros, so hit two, and it's going to ask us for a left bound. So we're going to go find this point here, so we want to go to the left, which is actually up here. The left most point that we see here, and we want a left arrow until we can get to where, oops, there it was. So here we are, we can press enter there, that's close enough, and then we want to go across the x-axis to get the right bound, press enter, enter for the guess, and there's our negative three. So we found that one. If we want to find the other one, again we can do second trace two for a zero, but I want to show you another way that you could find the zeros. It takes two equations, but you would put in zero here, y2 equals zero, because that's the x-axis, and then all you have to do is second trace five, and you would do the intersection. I want to get to that point over there, so I'm going to just go on my second equation because it's a straight line, and I just have to make sure I'm on the other side of the vertex. But there I am, so press enter, and then up here it says y1, yep, that's my other equation, enter, enter for the guess, and there we find x is equal to six. So if we had drawn our graph, we would have had it at negative three, and six, four, five, six, and our graph looked something like this, and that would be our graph solution. Square root property is another way that we can solve it, so you have x squared or something squared is going to be equal to a constant. So if that's true, that means that x is equal to the square root of k, or it could also be the opposite of square root of k. For example, before we do our problems, if I take two and I square that, I get that four, but if I take negative two and I square that, I also get four. So that's why you have to have the positive and the negative. Now if you remember, we talked about the fact that it was something squared. So what we're really doing here is taking the square root of both sides, this something squared is literally a something squared. It's a binomial squared, but a square root and a square cancel each other out. They're opposite functions, so that just leaves us with p plus five, and then we have plus or minus the square root of 49, which will be plus or minus seven. And then we just subtract the five from both sides, but we're going to have seven minus five and negative seven minus five. So p is going to be equal to seven minus five is two, or p will be negative seven minus negative five, which is negative 12. We have both of those answers. In this case, I do have just something that looks like x squared, but I have to take the constant to the other side. It has to be the something squared equal to a constant. So we want to take the square root of both sides as plus or minus the side. So square cancels the square root. So we have m equal to plus or minus the square root of 20. And I'm good with that. That's what we call an exact answer, because we haven't rounded it if we try to take the square root. And you don't have to simplify it any further. All right. So let's fill in the blank to make a perfect trinomial square. A perfect trinomial square is a perfect square. And the last term is also a perfect square. But when I take two times the product of that, I should get two thirds. What does that really mean? Well, let's think about it this way. If I take one third and I double that, I'm going to get two thirds. And if I unsquare p, I'll get p. So two times one third times p is equal to two thirds p. But this last term has to be a perfect square. So it's actually going to be p squared plus two thirds p plus one ninth. And then it says factor it as a binomial square. Well, binomial squared, you unsquare this term to get your p because it would be p times p. That's how you get your p squared. And it's a plus in the middle here. So we have a plus and you unsquare this. The square root of one ninth is equal to one third. And so we have one third here because we'd have plus one third plus one third p squared. One third p plus one third p would give us our two thirds p and one third times one third is our one ninth. So we have p plus one third quantity squared. So we want to complete the square here. So we're doing the same thing. We have p squared plus six p. And we want to add something like we did up here to make a perfect trinomial square. But if I add it to one side, I have to add it to the other side as well. Well, the really the trick here is to think of if I had taken two thirds up here and multiplied it by one half or taken one half of it, the twos would have canceled and I would have had one third. So this number that I have to square for the last term is half the middle. Or b over two. My b is six and I divide that by two and that gives me three. But I have to square that three. This one third quantity squared was b over two quantity squared. So six over two is three. Three squared is nine. So it's p squared plus six p plus nine is equal to I have to add nine to both sides. So negative four plus nine will be five equal to and I want to have this binomial squared here. Remember you unsquared the first term that's p and unsquared the last term that's three. Or half of this middle term will also be what you're going to add to positive here. So I want a positive here and now I'm ready to solve. So now I have a perfect square equal to a number just like we had before. So I take my square root of p plus three quantity squared and the square root cancels each other out p plus three. And I have plus or minus the square root of five which is simply plus or minus the square root of five. I can't go any further with that. It's not a perfect square. And if I subtract the three from both sides I have negative three plus or minus the square root of five. And the last way to do it is to use the quadratic formula. Well the quadratic formula is going to be this negative b plus or minus b squared minus four c over two a. I am not going to make you do that by hand. If we call up the calculator we have this program down here. Now you may not have it. Hopefully you've gotten it from me at this point but if you haven't next class period we will. You have the program and it should say quantity four. Yours may not work exactly like mine but they all give you the same answer. Before I can do the quadratic formula here I have to make this equal to zero. Three n squared minus seven and minus six equal to zero. And this is what I will put in my calculator for the quadratic formula. And a here is the coefficient on n squared so three. B is the coefficient on n so negative seven. And c is the constant or negative six. Again using my calculator I just choose program and then I press enter because I want quantity four. It says program quantity four. Yep that's what I want so another enter. And then I enter my a which was three enter. B which was negative seven enter. C which is negative six and enter. And I find out that x is equal to three and negative this is two thirds. So x is equal to three and negative two thirds.