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Robert McKeown: Hello and welcome to ALEKS

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walkthrough video number three. My name is

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Robert McKeown, and I'm very happy that you

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decided to check out my video. We're moving

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along now, we're still talking about some very

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fundamental algebra skills. But we're going to

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introduce two new concepts to you. The first is

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the idea of a quadratic equation, where we

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possibly have more than one solution. So that's

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going to come up today if it didn't come up

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before. And we're going to look at exponents and

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the rules for exponents. Now, students often

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tell me or when they get this part in the

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course, they say, Oh, you know, this is very

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easy. I kind of know this, or they see the rules

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like, Oh, yeah, yeah, I get it. Well, you have

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to know it inside and out. Because you're going

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to be given problems in economics, you could be

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given math problems in your math courses. And

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you're expected to be absolute masters of these

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topics. And you're going to have to do a whole

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bunch of different operations, and exponents are

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just going to be one of them. So it's really,

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really important to nail the rules now, how

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exponents can be manipulated, so that when you

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get more complicated problems, you don't make a

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silly mistake, and then end up getting the wrong

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answer. So per usual, I want you to get your

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pencil, I want you to get your paper, your

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booklet, you can print out the slides, you can.

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If you have an extra device, if you have a

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tablet, you can write your answers on the slides

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electronically and save them saving them as

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optional. But you should work through these

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problems with me work through them together.

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That's how you're going to master algebra.

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That's how you're going to master mathematics by

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practicing and doing yourself. Now, let's get to

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it. Let's take a look at our first question. The

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question is asking us to graph the solution to

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the following inequality. So what is it mean?

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Well, let's take a look at the slides. And let's

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go into some more detail. So there's the

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question written again. What is the solution to

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this inequality? Well, what it's really telling

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us and I'll show it to you, I'll type it out is

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find the x values, or the values for x? And

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which the equality is true? So remember, that if

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we're multiplying two positive numbers, the

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answer is also going to be positive. If we're

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multiplying a negative and a positive, the

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answer is negative. And two negatives,

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multiplied by each other gives us a positive Not

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to be confused with two wrongs making a right.

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So we need to find the values of x for which

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this thing is strictly greater than zero.

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Looking at the expression, I want you to notice

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two things. Notice that if x is equal to

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negative six, or maybe I should say if x is a

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number smaller than negative six,

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then x plus six will be less than zero. And if x

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is equal to five, or again, I'm going to change

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it from equal to

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less than five.

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x minus five will also be less than zero. Now in

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general, how do you solve this type of problem?

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Where we've got two factors and, and they're

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multiplied by each other? What are we going to

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do to solve this? We're going to create a sine

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diagram. And I had my students on the econ 1530

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final exam. Last year, my students have to draw

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sign diagrams to answer questions that are a

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little more complicated than this, but are the

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same idea. So I'm going to show you how to

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create a sign diagram. And if you're in my

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class, you're going to want to be able to do it,

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because I'm going to give you marks if you can

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do it, too. So how do we create a sine diagram?

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Well, I'm going to start off by writing the two

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factors over here.

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And then I'm going to draw, well, maybe I'll

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draw a line like that. And I'll have negative

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infinity over here. I'll draw a line over here.

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I guess I could put a matching line there if I

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wanted to. And I'm going to mark off these

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important numbers, the negative six and the

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five, they're going to be important. And zero

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might be important as well, or might not so but

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I'll put in the zero as well. Now, if x takes on

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a value, less than negative six,

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x plus six will be negative. And if x takes on a

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value less than five, x minus five, will be

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negative. Now, if x is greater than negative

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six,

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x plus six is going to be positive. And if x is

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greater than five, x minus five will be

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positive.

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And I'll just since I've added in those vertical

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lines, we'll add that negative symbol there now.

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two negatives. A negative times negative is a

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positive. A positive times a negative is a

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negative. And two positives, a positive times a

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positive gives us a positive. So we can see here

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that if x is less than negative six, this

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inequality will hold. And if x is greater than

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five, this inequality will also hold. And notice

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that if x is equal to five, it does not hold

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because the answer will be 00 times any real

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number is equal to zero. And similarly, if x is

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equal to negative six, this expression will be

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equal to zero, and therefore it is not strictly

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greater than zero. Now let's put our answer into

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ALEKS. I'm going to use the circle that sort of

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hollow. And I'm going to choose negative six,

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and five. And I want my lines to be

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where x is greater than five, and x is greater

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than negative six. Now let's see if we've got

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the right answer. And we did it successfully, we

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use the sign diagram to make sure that we got

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the right answer. So these questions are a

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little more, a little more time consuming, a

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little more challenging than you might think if

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you try and eyeball it, you might get the

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answer, but you also might make a mistake. So in

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this question, we have squared x, and it's equal

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to 50. It tells us that x is a real number. But

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what it wants us to do is to simplify our answer

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as much as possible. And I'll explain what the

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question means when it says that. In fact, you

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can just see at the top there, there's something

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about this square root property, so I'm going to

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show you the square root property. The other

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thing to be aware of is that it's possible that

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there is no solution to This question. And it's

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also possible that there's more than one

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solution.

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Looking at the slides, we have the same

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question. I've just added in that little hint.

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ALEKS really wants us to use the square root

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property here. As a first step, I want to

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isolate X on one side of the equation equation.

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Now I've got x squared on one side of the

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equation, but what I really want is x. So I'm

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going to take, well, I could just write it like

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that, I'm going to take the square root of both

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sides of the equation. But as I'm doing that,

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notice that if I had x squared is equal to 16.

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And I'll just use, you know, an easy number

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here, x could be equal to four, or x could be

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equal to negative four. So I'm taking the square

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root of both sides of the equation. But I want

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to recognize that x could be positive or

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negative. So I'm going to have x over here. And

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I'm going to say that's equal to plus minus the

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square root of 50. Now you probably notice and

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saying, Well, look, I've got x squared is equal

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to 50, I could just take the square root of 50.

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And I've got the numeric solution. That's not

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what ALEKS wants us to do once to test our

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algebra skills. There are reasons in the real

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world that we might not want to just solve away

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for this, when you get into 1540. And you're

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looking at multivariate situations, there's not

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going to be a just a simple solution, like the

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square root of 50 is equal to a number a little

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bigger than seven. So what can I do to further

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simplify this? Well, 50 I can do something with

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that 50. So what if I wrote 50 as 25 times two.

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And I know that x is equal to plus minus, what's

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the square root of 25, it's five. So I could

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bring the five out and leave the square root of

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two, which of course is going to be an

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irrational number. Now let's see how I did we're

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going to go back to ALEKS and we're going to

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plug in this solution, notice there's two

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solutions. So x could be equal to five times the

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square root of two or x could be equal to

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negative five times the square root of two. So

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here we are an ALEKS x is going to be equal to

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five. And I click on the square root symbol

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after I put in the five times the square root of

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two, and then I want another O. So I'm going to

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start off by clicking this little button here.

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And then I'm going to make sure that my cursor

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stays on the left hand side there. And then I

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want five times the square root of two. And then

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moving into the second box, I want negative five

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times the square root of two. And let's see if I

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did it right. And I did. So we have the right

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answer here. And there were two solutions. This

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little question that I've added to demonstrate

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the introduction to factoring. And so here the

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question, we have an expression, and we're being

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asked to expand it. So how are you going to

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expand it? Well, it's good to remember that the

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expansion of this thing is going to be one times

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three, plus one times four, plus two, times

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three, plus two, times four. So you do the outer

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first and then the inner. Expanding, we get x

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times x plus x times four, minus one times x

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minus one times four x squared plus four x minus

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x

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minus four. And we can collect the like terms.

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And we're left with x squared plus three x minus

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four. And this is known as a quadratic

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expression. It's an expression because it's not

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equal to anything. It's not a, it's not a

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function, right. But we'll see. We'll see what a

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function is later. Now let's look at a more

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complicated problem. So I haven't showed it to

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you and ALEKS will show you we'll put the answer

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into ALEKS later. But the questions asking us to

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factor the following completely. This is a

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tricky, factoring, question. So to start off,

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how are we going to find out what y is equal to

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that's another way. Another thing that we're

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trying to do here, when we are asked to factor

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it into its simplest multiplicative

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multiplicative bits? Well, as a first step, why

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don't I factor out why four, and so we're left

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with nine y squared minus 24, y plus 16. And you

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can see that this inner thing is a quadratic

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expression. Now, how can we solve and factor

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this quadratic? Well, one method would be to use

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the quadratic formula, you can look up the

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quadratic formula online. That would be one way

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that you could go about factoring this. But if I

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look at a little more closely, I see that I

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could rewrite this just a little bit. Oops.

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And this is kind of, as we'll see, these are

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equivalent. As we'll see in the next question, I

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can

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factor this out. Like so where you can see that

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if I multiplied this thing out, we'd have minus

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24, y in the middle 12 y minus 12 y, and the two

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ends are going to get us back to this thing up

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here. The trick is to see that both three and

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four are factors of 24. And when three times

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four, we get 1212 minus 12 gives us 24. That's

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sort of the secret. So let's see if it's the

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right answer on ALEKS. Here's the problem on

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ALEKS. So we had y and I'm going to press sift

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while my keyboard shift six and I can

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automatically get an exponent for I hit the

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right move key and I start open up a bracket

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and I'll do it I just realized I made a little

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mistake over here. I in my head. I knew I was

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doing it correctly, but I should have had a

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00:19:10,230 --> 00:19:12,870
negative sign over there. Now going back to

250
00:19:12,870 --> 00:19:16,920
ALEKS. Here we are. I'm going to click the check

251
00:19:16,920 --> 00:19:22,500
button and we got the right answer.