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Contents

## Pick a Point

We've now found out that if Grant sets his price per wiper blade set equal to \$46, he will earn the maximum profit he can of \$290,000. But, of course, there may be factors that prevent him from setting his price exactly equal to this amount. And he wants to rule out a few other prices that would lead to the worst case scenario, or at least one of the worst case scenarios, no profit at all. If Grant wants to know what he would need to set his price per wiper blade set equal to, to have his profit equal 0, what point or points on our graph are we interested in? Do we want to know about the vertex, the maximum, the minimum, the c-intercept or intercepts, or the p-intercept?

## Pick a Point

If Grant is making no profit, that means that p equals 0. The points where p equals 0 are the c-intercepts. So those are what we're interested in.

## Can we factor

The next question is whether or not we can find these c-intercepts simply by factoring. By factoring here, I mean the traditional sense of factoring using that first method we discussed pretty early on in the class.

## Can we factor

The answer is, no, we can't. I can't think of two numbers that add to equal even if those numbers do exist, they're going to be really, really hard for me to think of. So what are we going to do now? We can't just read these off the graph either. It's really hard to tell exactly where they are.

## Equation for x intercepts

We're starting to see that some equations for parabolas can't be factored. When that's the case, we need another method of finding where they intersect the horizontal axis. Let's look at this equation, y equals 3 x squared plus 6 x minus 10. Now if we want to find the x intercepts, what equation are we really trying to solve? Please put you answer with the full equation in this orange box.

## Equation for x intercepts

We just need to solve 0 equals 3 x squared plus 6 x minus 10. Since y equals 0 at the x intercepts.

## Complete the Square

We actually worked through a problem pretty similar to this at the end of the previous lesson. It was a different equation, and the leading coefficient was 1, so things were a little bit simpler, but the same sort of technique that we used there applies here. I'd like you to complete the square, with the expression on the left side of the equation, and to get this equation to the point where there's a non-zero integer on the right-hand side, and a term with our perfect square in it on the left hand side.

## Complete the Square

To complete in this square, we end up with the equation 3 times the quantity x plus 1 squared, equals 13.

## Get rid of the 3

Remember that we're trying to get x by itself. So the next step we need to take in order to do that is to get rid of this factor of 3 out in front here, so that I can rewrite the equation with a factor of 3 gone from this side. Of course we need to do this legally, so think about what that means for the right side of the equation as well.

## Get rid of the 3

This leaves us with the quantity x plus 1 squared, equals 13 over 3.

## What is the next step

Remember that we're trying to get X by itself. So what's the next step that we should take to move towards that? Should we square both sides of our equation? Should we find the positive square root of each side? Should we find the negative square root of each side? Or should we find both the positive and the negative square roots of both sides?

## What is the next step

The last choice is correct. We need both the positive and the negative square roots. This time, we don't have any real life situations ruling out negative numbers, and it makes sense mathematically that if we end up with a negative number inside the parentheses on the left-hand side, since we're squaring it, it will equal a positive number.

## Plus or Minus

Now, to indicate that we want the positive and the negative versions of a number, we can write this symbol out in front, plus or minus. Now, in MathQuill, to create this symbol, you need to write backslash pm and then a space. The next step, like we said, is going to be to take the square root of both sides. Including both the positive and the negative routes. I've already dealt with the left side for you and I'd like you to fill in the right side.

## Plus or Minus

We have x plus 1 equals plus or minus the square root of 13 over 3. Now, you may we wondering why it doesn't look like I took the positive and negative versions of the square root of the left side of the equation. But with this equation right here, we're actually covering all of our possible solutions for x. Here's why. We know that if we have a square root of a number, and we make that negative. Then that's just equal to negative 1 times the square root of that number. The negative version of this square root effects would then be negative equation as well, with the same thing on the right side. They're more effectively taking care of the plus and minus versions of the left root. This is going to give us four different equations, which are these four equations right here. However, we need to notice something special about these equations. If we take this equation, negative x minus 1 equals negative root 13 over 3, and we divide both sides by negative 1. We end up with x plus 1 equals the root of 13 over 3, which is just this equation up here. That means that one of our equations in the second pair is the same as an equation in the first pair. Now if we do the same thing for these 2 equations. The 2 we haven't touched on yet. We'd find that they are equivalent as well. So if we take the positive and negative square roots of both sides of the equation, So it saves us some time to only put the plus or minus in front of 1 side of the equation. And just take the positive square root of the other side. Or the negative 1 if you prefer that.

## What is x

Now all that's keeping x from being by itself is this plus 1. So move that to the other side and create an expression for x. Remember our handy dandy plus minus sign.

## What is x

This gives us x equals negative 1 plus or minus the square root of 13 over 3.

## Separate Solutions

Remember that this plus or minus symbol actually indicates that we're going to have two separate numbers here. There are two numbers contained in this line right now. One that's equal to negative 1 plus this quantity, and one that's equal to negative 1 minus this quantity. What I'd like you to do now is to find the two values of x that are produced by this. Remember what this symbol means to get you those two different answers, and when you find them, round each of them to two decimal places. Please put the answer that is the lower number in the top box, and the answer that is higher number in the bottom box.

## Separate Solution

The two answers we get are negative 3.08 and 1.08. Here's how I found them. I found the first one by using a subtraction sign in here. So this is negative 1 minus the square root of 13 over 3. And I found the second answer 1.08 by using the plus sign in here, negative 1 plus the root of 13 over 3. This means that our X intercepts are at -3.08, 0 and 1.08, 0. I would like to make one little change on here though. Since these are rounded answers, let's put approximately equal to signs between these. These are approximate numbers, and if we wanted to, we could continue calculating out these infinitely long decimals by looking for their out and calculating these.

## Generic Grouping

So, we eventually got to find the x-intercepts of this equation that we didn't know how to factor, but this is a pretty long winded process. Unfortunately, this is a situation that we actually run into quite a bit. It's really only very special cases of quadratic equations that we can factor. Thankfully, there's a formula that we can use to calculate the roots of a quadratic equation so we don't have to go through this long process every single time. Now, get excited, because you are going to work out that formula yourself. We're going to start with the general form of a quadratic equation, ax squared plus bx plus c equals complicated at certain points but once you've gotten to the end and you understand why the formula we end up with works, you'll get to use this every single time. Let's follow this same method as we did in the last example. First things first, let's complete the square. It might be a little bit messy but we're going in. The first thing I'd like you to do is to factor out a from these first two terms here. What belongs inside these parenthesis then?

## Generic Grouping

If we take out the factor of a from these two terms, the coefficient of the first term becomes one, and the coefficient of the second term becomes b over a. So we have a times the quantity x squared plus b over a x, and then, of course, plus c equals 0. Great.

Let's keep moving forward with completing this square. What terms belong in these 2 boxes to help us complete the square? Since the right side of the equation doesn't change from this step to this step, that means that the total value of the left side shouldn't change either. Think about what that means about the relationship between what belongs in this box and what belongs in this box.

The term that belongs in this slot is b over 2 a. We get that by dividing this middle coefficient by 2. Since we know that this plus itself has to equal b over a. Now what term belongs here? Well, this needs to compensate for the amount that we're adding to the entire expression on the left-hand side. When we inserted b over 2a right here, this is just adding a constant that is b over 2a squared times that coefficient of a that's outside. This simplifies to b squared over 4a. So that's what belongs right here.

## c and a leave the left

Let's lay back to what we did before. And what our goal is, eventually, we want to get x bytes off. We want to isolate it on one side of the equation, so that means that something is going to have to happen with all this constant stuff over here. Something is also going to have to happen with our coefficient of a. So actually, let's do with both of those things now. What does that will leave us with on the left side of the equation and what does that put on the right side?

## c and a leave the left

So, you could either choose to subtract the c or divide by a first. It doesn't matter which order you do these in, as long as you obey all the proper mathematical rules. In particular, when we divide both sides by a, make sure that you divide every term on each side by a, and do that division correctly. The first time I did this, I actually messed up dealing with the second term in the left-hand side. I didn't write a squared, I just didn't write an a at all in the denominator. So, everyone makes mistakes. It's not a big deal if you do. We just gotta keep learning from them. Anyway, the equation we end up with at the end of this step is the quantity x plus b over 2a squared minus b squared over

## What belongs on the right side

Now I'd like you to rearrange things, so that the only thing we have left on the left side of the equation, is x plus b over 2a squared. What belongs on the right side then?

## What belongs on the right side

When we have this by itself in the left hand side, the right hand side reads b squared over 4a square minus c over a. We got this by adding b square over 4a squared to both sides.

## Numerator and Denominator

We have 2 fractions on the right hand side of the equation right now. So Id like us to combine them. Think about the least common multiple of 4a squared and a is. And that will be the denominator. And then think about what the enumerator will become based on that.

## Numerator and Denominator

The least common multiple of A and 4a squared is 4a squared. To make the denominator in the second fraction equal that, we need to multiply both the top and the bottom fraction by 4a. Remember, this is the same as multiplying by 1, so it's totally kosher to do within an equation. Written as a single fraction then, this is B squared minus 4ac over 4a squared.

## Take the Root

We're so close to having x by itself. If I want the left side of the equation to now read x plus b over two a, what does the right side need to be? I'll make the version of the equation we just had a little bit bigger. Read each of these answer choices really carefully to figure out whether or not it's correct. Note that I've put square boxes next to each of these. Indicating that you can check as many of these as you think are right. There might be multiple ways to write what belongs on the right side.

## Take the Root

We want to take the square root of both sides of the equation. We want to make sure that at least one of those sides has a plus or minus next to it. Since the left side doesn't, the right side needs it. So that rules out all of our answers that don't have plus or minus symbols in front of them. This first answer is definitely correct because it has a square root sign over the entire expression, those on the right hand side originally. However, when we have a square root over a fraction. This is the same as taking the square root of the numerator over the square root of the denominator. So this choice is correct as well. If we simplify that denominator, square root of 4a squared, that just gives us 2a, so the last choice is also correct. There's so many different ways of writing this, however I think that this last one is probably the most simple, since it has the fewest radicals involved.

## One Fraction

Finally we're about ready to get x by itself. All we need to do is get rid of this constant term, b over 2a, that's right now on the left side of the equation. I'd like you to get x by itself and then I'd like you to write our final answer on the right hand side as a single fraction.

## One Fraction

To get x by itself, we need to subtract b over 2a from both sides. Then, both of our fractions on the right-hand side have a denominator of 2a, so it's really easy to combine them. This gives us, finally, a final answer of x equals negative b, plus or minus the square root of b squared minus 4ac all over 2a. You can heave a giant sigh of relief now, that was a lot of work and it involved some complicated Algebra. I think that in doing this, you had to use almost all of the skills you've learned in this course so far. So, this is super impressive, I am so proud of you. So, what does this mean again? Remember, we started out with an equation in this form, a quadratic equation, ax squared plus bx plus c equals 0. Working from that, we've solved for x which means that now, we have a formula where if we know the coefficients of a quadratic equation written in this form, we can find the values of x that satisfy it. In other words, for any parabola we have, if we can write its equation in standard form, ax squared plus bx plus c equals y, then we can set y equal to 0 and find the x-intercepts. This is huge, this is incredibly powerful. And we did all of these by just completing the square. And, of course, doing some other algebraic manipulations.

This formula we just created has a special name. It is called the quadratic formula. And this name makes sense. Remember, this gives us solutions for quadratic equations, which are written in the form ax squared plus bx plus c equal 0. As long as we can find these coefficients, we can solve for x. So let's try this out. Let's see if it really works. Let's look at an example we did before, and see if we can get the same answers for the x coordinates of its x intercepts, as we did using other methods. Let's look back at x squared plus 5x plus 6 equals 0. For this quadratic equation what are a, b, and c? Were these in reference to the general form of the quadratic equation that we've been talking about?

A equals 1, b equals 5 and c equals 6. Since we should have ax squared plus bx plus c. Equals 0, of course.

## Numbers in Boxes

Here are the values for a, b and c that we found in the last quiz, and now I'd like you to use these values in the quadratic formula. On your own piece of paper, please write in these values in the proper places and then simplify, so that you can write a single number in each of these boxes right here.

## Numbers in Boxes

So an initial plugging in of these numbers gives us this right here, which I'm not going to read out because as you know, I'm pretty lazy. Anyway, now I'll go through and simplify within each of these spots to figure out what numbers belong there. I'll leave that negative 5 out front. Under the radical we'll have

## 2 values of x

Now that we have our numbers plugged in and simplified in our quadratic formula. Tell me what two values of x satisfy our original equation. So come up with two, just numbers that belong in these boxes. Please put the smaller one here and the bigger one here.

## 2 values of x

Our plus or minus here means that we can either have negative 5 plus the root of is just 1. So this number should be negative 2 and this number should be negative 3.

## Values of a b c

In the last quiz, we got 2 values of x that satisfy this equation, which were the same as the ones we found through factoring earlier on. It's definitely quicker to factor if you can, but sometimes it's hard to see right away how to factor an expression, so if you don't know how, now you have another way to solve for x in situations like this. Let's look at the equation 4x squared minus quadratic formula to solve it, even if there are other methods of doing it. So what values of a, b and c should we use for this equation, if we are going to use the quadratic formula?

## Values of a b c

We'll have a equals 4, b equals negative 4 and c equals negative 3.

Now that we have a, b, and c picked out, use these in the quadratic formula to solve for x.

Using the quadratic formula, we get values of x of 3 halves and negative 1 half.

## Price for no profit

And now finally back to Grant. Here's his equation for profit written in the standard form for an equation for a problem. And if you remember back at the beginning of the lesson, he wanted to know which prices for wiper sets will cause him to not make any profit at all, or to make p equal 0. I'll tell you right off the bat that there are 2 of them and I'd like you to round each of them to 2 decimal places.

## What is a root

I think it's time that we introduce a pretty important word, the root of a polynomial. This is a word we're going to hear more and more in the coming lessons. So I think we should get used to it now. So if the root of a polynomial is a number n, such that when you substitute n in the spot of x in that polynomial, it makes the polynomial equal to 0. Then what do you think the root is? Where do we get that number that is the root from? Is the root just 0? Is it the x coordinate of the vertex of the parabola? Is it the y coordinate of the vertex? Is it the x coordinate of the y intercept? The y coordinate of the y intercept? The x coordinate of the x intercept? Or the y coordinate of the x intercept?

## What is a root

A root of a polynomial is the x coordinate of one of its x-intercepts. We've been looking at equations for parabolas that can be written in the form y equals ax squared plus bx plus c. And then we want to find the x intercepts, we replace y with 0. And come up with a quadratic equation. Any numbers that satisfy this equation, are roots. Well we've been solving for x using factoring, or the quadratic formula, or finding the roots of these quadratics. It's convenient to have the word root, because I've been asking you for x intercept over, and over again. But for every x intercept, the y value is 0. So just keep in mind that a Y value that goes along with the root to create the coordinates of a point is 0. Just like we plugged in here.

## Find all the roots

Now that I know our root is the x coordinate of the x-intercept, I'd like you to find all the roots of the y equals to 4x squared plus 4x plus 1. Please put all of your answers in this box but separate them with commas. Keep in mind what this means in terms of our graph.

## Find all the roots

If we're looking for the roots of this, then we need to set y equal to 0, since the roots are the x-values that will satisfy the quadratic equation that goes along with this. Now, I could factor this, that would be one way of solving, but I'm going to use the quadratic formula just to see what happens. So, I have negative 4 plus or minus the square root of 4 squared minus 4 times 4 times 1, all over 2 times 4. Let's simplify. Now, this is starting to look pretty interesting. We have 16 minus 16 under the radical sign. So, that means that we're adding 0 or subtracting 0 from negative 4. This is the only number that we can add and subtract negative 4 and get the same answer, just negative 4. This is pretty strange, it's quite the conundrum. If x equals negative 4 over 8, that's a single answer. So, we get a final answer that the root is negative 1 half, and there's only one of them.

## Only one root

So we just found only one root for the polynomial 4x squared plus 4x plus 1. My question is what does only having one real root mean for the graph of the equation? Does it there's only on x-intercept? Does it mean there is only on y-intercept? Does it mean that there is only one vertex, or does it mean that we made a mistake?

## Only one root

It means that there's only one x-intercept. To see this in action, let's take a look at the graph of the problem that we were looking at on the last quiz. So, here's the graph of y equals 4x squared plus 4x plus 1. We found before that the root of this was negative 1 half. And we can see that the x -intercept is at negative 1 half, 0, just like we would expect. This parabola is special and that instead of crossing the x-axis going down and then crossing again, in other words, having 2 x-intercepts, it simply touches the x-axis at a single point, which is its x-intercept. Another example of a parabola that does this is the basic one, the parent graph, y equals x squared. You can imagine if we shifted this over and made a little bit wider, we would get back to this graph.

## Getting to the root of the problem

Let's take a look at one last problem before we finish this lesson. I'd like you to find the roots of the parabola y equals x squared plus 2x plus 10. I'd like you to use the quadratic formula to do this, and get your values for x in the form, so that there's just a single number in each of these boxes .

## Getting to the root of the problem

Since we're trying to find the roots of this equation, we need to set y equal to where y equals 0. Now, let's use the quadratic formula to solve this. Plugging the proper coefficients into the proper spots in the quadratic formula, gives us x equals negative 2 plus or minus the square root of 2 squared minus 4 times 1 times 10, all over 2. Let's simplify. Well, this is getting pretty interesting. Look at what I have underneath my square root sign. We end up with negative 2 plus or minus the square root of negative 36 over 2. This is pretty shocking. This problem is asking us to take the square root of a negative number. We can't do that. We do not have any idea what real number this would equal. In fact, there is no real number that this equals. So, what does this mean? No solutions? No crossing in the x-axis? Well, we found a solution. Let's take a peek at the graph. So, here's the graph of y equals x squared plus 2x plus 10. Remember, when we were looking for roots, we're looking for the x-coordinates of x-intercepts of the graph. But look at were our vertex is, it's right here way above the x-axis and our parabola is point upward. Coming to the y-coordinate at this point right here is the smallest y-coordinate of any point on this graph. In other words, because this is the minimum point, the graph is never going to get any closer to the x-axis, and so it's never going to touch it at all. This is pretty fascinating. The weird thing is that we still got some sort of answer. You'll have to wait for the next lesson though, to find out what this means.

## 1

Now that we have our numbers plugged in and simplified in our quadratic formula. Tell me what two values of x satisfy our original equation. So come up with two, just numbers that belong in these boxes. Please put the smaller one here and the bigger one here.

## 1

Our plus or minus here means that we can either have negative 5 plus the root of is just 1. So this number should be negative 2 and this number should be negative 3.