Introduction to the quadratic equation Video

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Introduction to the quadratic equation

Welcome to the presentation on using the quadratic equation. So, the quadratic equation it sounds like something very complicated and when you actually first see the quadratic equation, you’ll say, “Well, not only does it sound like something complicated but it is something complicated.” But hopefully you’ll see over the course in this presentation that is actually not hard to use and then in the future presentation I’ll actually show you how it was derived. So, in general you’ve already learned how to factor a second degree equation. You’ve learned that if I had to say x2-x, let’s make that a minus, x2-x-6=0. If I have this equation, x2-x-6=0 that you could factor that as (x-3)(x+2)=0 that means that x-3=0 or x+2=0. So, x=3 or -2 and a graphical representation of this would be, if I had the function f(x) is equal to x2-x-6=0. So, this axis is the f(x) axis. You might be more familiar with the y axis and for the purpose of this type of problem, it doesn’t matter and this is the x axis. If I were to graph this equation, x2-x-6=0, it looks something like this. This is f(x)=-6 and the graph will kind of do something like this. I know it goes through -6 because when x=0, f(x) is equal to -6. So, now it goes with this point and I know that when f(x)=0, so f(x)=0 along the x axis, right because this is one, this is zero and this is negative one. So, this is where f(x) is equal to zero along this x axis, right. We know that it equals to zero at the points x=3 and x=-2. That’s actually what we solve here. Maybe when we were doing the factoring problems you didn’t realize kind of graphically what we were doing. But if we said that f(x) is equal to this function and we’re setting that equal to zero. So, we’re saying that this function went as this function equal to zero. When is it equal to zero? Well, it’s equal to zero at these points, right because this is where f(x)=0 and then what we were doing when we solve this by factoring is we figured out the x values that made of f(x)=0 which is these two points and just a little terminology. These are also called the zeros or the roots of f(x). Let’s review that a little bit. So, if I had something like f(x)=x2+4x+4 and I ask you where are the zeros or the roots of f(x)? That’s the same thing as saying where does f(x) intersects the x axis? It intersects the x axis when f(x)=0, right. If you think about that graph I had just drawn. So, let’s say if f(x)=0, then we could just say 0=x2+4x+4 and we know we can just factor that and that’s (x+2)(x+4) and we know that zero is equal to zero at x=-2. Well, that’s x=-2. So, now we know how to find the zeros when the actual equation is easy to factor. But let’s do a situation where the equation is actually not so easy to factor. Let’s say that we had f(x)=-10x2-9x+1. When I look at this even if I were divide it by 10, I would get some fractions here. It’s very hard to imagine factoring this quadratic and that is what I actually call the quadratic equation or this second degree polynomial. So, we are trying to solve this because we want to find out when it equal to zero, -10x2-9x-1. We want to find that what x values make this equation equal to zero and here we can use a tool called the quadratic equation. Now I’m going to give one of the few things in math that’s probably a good idea to memorize. The quadratic equation says that the roots of a quadratic are equal to and let’s say that the quadratic equation is ax2+bx+c=0. So, in this example a=-10, b=-9 and c=1. The formula is the roots x=-b+/– √b2-4ac/2a. I know that looks complicated but the more you use it you will see it’s actually not that bad and this is a good idea to memorize. So, let’s apply the quadratic equation to this equation that we just wrote down. So, I just said and look a is just the coefficient on the x2 term and b is just the coefficient on the x term and c is a constant. So, let’s apply it to this equation and what’s b? Well, b is negative nine, right and we can see here, b=-9, a=-10, and c=1, right. So, if b is negative nine, so let’s write that’s -9 +/-√-92. Well, that’s 81 -4 x a, a=-10 x c which is one. I know this is messy but hopefully you're understanding it and all of that over 2a. Well, a=-10, so 2a=-20, right. So, let simplify that, -9 x -9 = 9 +/-√81. We have a -4 x -10 and this is a -10. I know it’s very messy. I really apologize for that x 1 so, -4 x -10 = 40 and then we have all of that over negative 20. Well, 81+40=121 so this is 9 =/-√121/-20. The square root of 121 is eleven, so I’ll go here and hopefully that you don’t lose track of what I’m doing. So, this is 9=/-11/-20 and so if we said 9 + 11/-20 that is 9 +11=20 so, it’s 20/-20 which equals to negative one. So, that is one root that’s nine plus because this is plus or a minus and then the other root would be 9 -11/-20 which equals to minus two over minus twenty which is equals to one over ten. So, that’s the other root. So, if we were to graph this equation we would see that it actually intersects the x axis or f(x)=0 at the point x=-1 and x=1/10. I’m going to do a lot more examples in part two because I think if anything it might have just confused you with this one. So, I’ll see you in the part two of using the quadratic equation.

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Welcome to the presentation on using the quadratic equation. So, the quadratic equation it sounds like something very complicated and when you actually first see the quadratic equation, you’ll say, “Well, not only does it sound like something complicated but it is something complicated.” But hopefully you’ll see over the course in this presentation that is actually not hard to use and then in the future presentation I’ll actually show you how it was derived.

So, in general you’ve already learned how to factor a second degree equation. You’ve learned that if I had to say x2-x, let’s make that a minus, x2-x-6=0. If I have this equation, x2-x-6=0 that you could factor that as (x-3)(x+2)=0 that means that x-3=0 or x+2=0. So, x=3 or -2 and a graphical representation of this would be, if I had the function f(x) is equal to x2-x-6=0. So, this axis is the f(x) axis. You might be more familiar with the y axis and for the purpose of this type of problem, it doesn’t matter and this is the x axis. If I were to graph this equation, x2-x-6=0, it looks something like this. This is f(x)=-6 and the graph will kind of do something like this.

I know it goes through -6 because when x=0, f(x) is equal to -6. So, now it goes with this point and I know that when f(x)=0, so f(x)=0 along the x axis, right because this is one, this is zero and this is negative one. So, this is where f(x) is equal to zero along this x axis, right. We know that it equals to zero at the points x=3 and x=-2. That’s actually what we solve here. Maybe when we were doing the factoring problems you didn’t realize kind of graphically what we were doing. But if we said that f(x) is equal to this function and we’re setting that equal to zero. So, we’re saying that this function went as this function equal to zero. When is it equal to zero?

Well, it’s equal to zero at these points, right because this is where f(x)=0 and then what we were doing when we solve this by factoring is we figured out the x values that made of f(x)=0 which is these two points and just a little terminology. These are also called the zeros or the roots of f(x). Let’s review that a little bit.

So, if I had something like f(x)=x2+4x+4 and I ask you where are the zeros or the roots of f(x)? That’s the same thing as saying where does f(x) intersects the x axis? It intersects the x axis when f(x)=0, right. If you think about that graph I had just drawn. So, let’s say if f(x)=0, then we could just say 0=x2+4x+4 and we know we can just factor that and that’s (x+2)(x+4) and we know that zero is equal to zero at x=-2.

Well, that’s x=-2. So, now we know how to find the zeros when the actual equation is easy to factor. But let’s do a situation where the equation is actually not so easy to factor. Let’s say that we had f(x)=-10x2-9x+1. When I look at this even if I were divide it by 10, I would get some fractions here. It’s very hard to imagine factoring this quadratic and that is what I actually call the quadratic equation or this second degree polynomial. So, we are trying to solve this because we want to find out when it equal to zero, -10x2-9x-1.

We want to find that what x values make this equation equal to zero and here we can use a tool called the quadratic equation. Now I’m going to give one of the few things in math that’s probably a good idea to memorize. The quadratic equation says that the roots of a quadratic are equal to and let’s say that the quadratic equation is ax2+bx+c=0. So, in this example a=-10, b=-9 and c=1. The formula is the roots x=-b+/– √b2-4ac/2a. I know that looks complicated but the more you use it you will see it’s actually not that bad and this is a good idea to memorize.

So, let’s apply the quadratic equation to this equation that we just wrote down. So, I just said and look a is just the coefficient on the x2 term and b is just the coefficient on the x term and c is a constant. So, let’s apply it to this equation and what’s b? Well, b is negative nine, right and we can see here, b=-9, a=-10, and c=1, right.

So, if b is negative nine, so let’s write that’s -9 +/-√-92. Well, that’s 81 -4 x a, a=-10 x c which is one. I know this is messy but hopefully you're understanding it and all of that over 2a. Well, a=-10, so 2a=-20, right. So, let simplify that, -9 x -9 = 9 +/-√81. We have a -4 x -10 and this is a -10. I know it’s very messy. I really apologize for that x 1 so, -4 x -10 = 40 and then we have all of that over negative 20.

Well, 81+40=121 so this is 9 =/-√121/-20. The square root of 121 is eleven, so I’ll go here and hopefully that you don’t lose track of what I’m doing. So, this is 9=/-11/-20 and so if we said 9 + 11/-20 that is 9 +11=20 so, it’s 20/-20 which equals to negative one. So, that is one root that’s nine plus because this is plus or a minus and then the other root would be 9 -11/-20 which equals to minus two over minus twenty which is equals to one over ten.

So, that’s the other root. So, if we were to graph this equation we would see that it actually intersects the x axis or f(x)=0 at the point x=-1 and x=1/10. I’m going to do a lot more examples in part two because I think if anything it might have just confused you with this one. So, I’ll see you in the part two of using the quadratic equation.

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