WEBVTT mathematics/algebra-1/smith
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Welcome back to www.educator.com.
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In this lesson we are going to continue on with our solving of equations by looking at some other solving techniques.
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Some of these techniques that we have seen before is of course using factoring to solve quadratic equation.
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What we are going to add to that is the square root principle, how to complete a square and everyone’s favorite, the quadratic formula.
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One of the first ways that you can solve the quadratic equation is break it down into its linear factors.
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When you do so, you can use the principle of 0 products.
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We saw that the way that it works is we treat each of the factors as a number
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and we know that one or possibly both of those factors must be equal to 0.
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We borrowed each of them and set them equal to 0.
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Then we solved each one individually and we got our solutions.
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Now of course one downside to using a process like this is that it relies on our factoring techniques.
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We have to be able to factor one of those quadratics and be able to grab them and set them equal to 0.
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The reason why that is so much of a problem is because not all quadratics factor easily.
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In fact, some of it will factor at all for searching around.
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Let us take this one like 2x² – 1 =3x.
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Suppose we want to solve that quadratic equation, one of the first things we might do for trying to use that factoring technique
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is we will go ahead and get everything over to one side.
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Now that in itself is not too bad and then maybe we will attack this using something like reverse foil.
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We only have one option to get 2x², we must use a 2x and x.
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There is only one way to get -1, we have to use 1 and -1.
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The problem with this is if you go through and you check your outside and inside terms, here is what you get, -2x and 1x.
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When those are combine it will give you -x, that is not the same as our middle term.
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We know that factorization is not correct.
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You might be saying to yourself we just made a small error.
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Maybe this needs to be a -1 and the other one needs to be 1.
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Well, let us try that one out and see if it turns out any better.
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On the outside we have 2x, inside -1x, those combine to be 1x and unfortunately still not the same as our middle term.
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There are no other possibilities for rearranging our 2 and 1, but those are the only possibilities that I could use my factoring process.
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If I'm not able to factor this then how we are suppose to be able to find solutions.
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After all, we still want to be able to find solutions for even quadratics like this.
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For this reason alone, that we have to pick up some new methods for solving quadratics other than just the factoring process.
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Let us first look at a nice one and see the method we can use on that type of quadratic.
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In this type of quadratic, we only have one copy of x that is in the quadratic equation.
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You do not have another x, just one copy of x².
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In these ones, the good news is you can solve directly.
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This uses the principle of square roots.
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What that principal says, is that if you have something like an x² equals to a number then you can split this into 2.
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Just put it into x equals the square root of that number and x equals the negative square root of that number.
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What you are doing in practice is like you are taking the square root of both sides of your equation.
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The reason why you are getting one being positive and one being negative is the square here erases information of what the sign used to be.
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Then we will go ahead and give the ± to the other side.
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x could have equal to 8, but maybe it was equal to -8.
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It so this was a kind of nice if you can recognize that there is only a single copy of x running around and it is being squared.
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For more complicated polynomials we will have to use a very special technique.
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This one is known as completing the square.
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What completing square does is it puts it into a new form and we can simply apply the principles of square roots to that new form.
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The process of completing the square is a little bit lengthy, so let me walk you through how the completing the square works.
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The first thing that you want to do is isolate your variables to one side so get all your x’s on one side.
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If it does not have a variable, go ahead and put it on the other side of the equation for now.
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You want to make sure that the coefficient on your x² term is 1.
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If it is anything other than 1 then go ahead and divide everything by that number.
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We will be looking for a number that we can add to both sides of the equation so that it will go ahead and factor nicely.
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There is a great technique for finding this number.
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You take the coefficient on your x term, you divided it by 2 and you square it.
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After you have added that number, then you can simply factor.
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In fact, it should factor nicely at that point.
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It should factor nicely.
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If it does not factor nicely then double check your steps up to that one to make sure you follow them correctly.
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We have gotten to that point, now we can use the principles of square roots.
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This will be the square root of both sides and make sure that one of them is positive and one of them is negative.
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Then we will solve directly for our variable.
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This is quite a lengthy process using complete the square, but if you methodically go through all of the steps, it does get the job done.
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The reason why I like completing the square so much is because it will work on any quadratic equation.
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We do not have to worry if it is a special form, or if it is easily factorable, we can just use completing the square and it will work.
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Of course there is one major downside to using a completing a square.
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There are lots of necessary steps and it makes it very easy to make a mistake when trying to follow it.
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The good news is all of these steps can be packaged up into a handy formula that we do not have quite as many steps to memorize.
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This formula is known as the quadratic formula.
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That is definitely something you want to know to be able to tackle many more difficult types of quadratics.
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What the quadratic formula says is that if you are looking at a quadratic equation of the form ax² + bx + c = 0, its solutions are given by this.
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x = b ±√(b² )- 4ac ÷2a.
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It is quite a lot to keep track of but this is probably good one to memorize.
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There is some good news, even though it is not the prettiest looking formula, it is built directly from completing the square.
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Completing the square works on any quadratics so you may also use this on any quadratic equation.
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It makes it extremely powerful for solving a lot of those quadratic equations.
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Now that we have some more tools under our belt, let us take a look at using some of these new techniques.
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To determine what type of solutions that these many techniques will produce, we can use what is called the discriminate to help us out.
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The discriminate is actually just a part of the quadratic formula.
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It is the part that sits underneath the square root.
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The reason why that will tell us what type of solutions we get
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is because depending on what type of numbers underneath the square root it could be real or it could be imaginary.
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It could be rational or irrational.
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Let us see how we can break that down.
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Before looking at the part underneath the square root, that is the discriminate and say it is equal to 0.
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What that tells us is that we will get exactly one real number solution.
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If we are looking at the discriminate and it happens to be greater than 0, think of some positive number,
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then we know we will have two different real number solutions.
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This one is special because you can take it a little bit farther.
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If that number is not only positive, but it is also a square number, your solutions will be rational.
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If it is not a square number then it will be irrational.
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And we just have one more case to worry about.
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What happens if the discriminate is negative?
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In that case you will have imaginary number solutions.
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Think of earlier when we are dealing with a negative sign underneath the square root, these happen to be complex conjugates of each other,
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but I will explain more about these in another lesson.
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With so many new techniques to solve quadratics, how do you know what to use when just faced with any quadratic?
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After all, some work for only certain quadratics if they are factorable and some techniques will work on any quadratic.
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Here are some good tips on when you should use these.
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The factoring technique that we picked up at the very beginning is one of the fastest methods you could use.
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This means if you are looking at a quadratic and you can see that it is factorable
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then go ahead and use that method because you will save yourself a lot of time.
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If you end up struggling with it for quite a bit then maybe another technique is what should be used.
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You can use the principle of square roots if you only have a single copy of x².
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This is a nice method because it is a very straightforward method that does not involve factoring and formulas.
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You just solve directly.
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If you are essentially stuck for too long and you know those methods worked out then you could try completing the square.
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The good news is, it should be able to find your solutions, even though it does have quite a bit of steps.
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In practice we do not usually use that one since the quadratic formula is built from the completing a square
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and it will also work for any quadratic equation.
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The reason why I like the quadratic formula it is a little bit easier to work with
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and then we simply have to plugged everything into the formula and use it.
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Make sure you identify your A, B and C terms and know that this one is faster than completing the square.
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Let us try some of these techniques and see how it does.
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We want to solve this first one using the square root principle.
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To figure out why we even use that principle, notice how there is only one x and it is being squared.
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There is no need to factor this one, we will just go ahead and work on getting x all by itself.
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Add 5 to both sides and then we will divide both sides by 2.
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And then we will take the square root of both sides.
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This is where that property comes into play because I have x² equals a number.
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I either have x = √5 or x = √(-5).
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Let us get some more space in there.
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Just solve for this one directly.
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Let us try the method of completing the square with this one.
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2x² + 4x – 2.
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There are lots of steps involved so carefully watch how I walk through this one.
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The very first thing I want to do is just start sorting things out.
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I will get everything associated with a variable on one side, if it does not have a variable I will move it to the other side.
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I'm adding 2 to both sides I do not have to worry about that 2 just yet.
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I want to check the leading coefficient to make sure that it is a 1.
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Unfortunately it looks like this one was not 1.
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To make it a 1 I’m going to divide everything by 2.
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2x² ÷ 2 = x², 4x ÷ 2= 2x and 2 ÷ 2 =1.
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Now comes the part where I need to add something to this quadratic, so that it factors nicely.
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The way we find out what this number is, is we look at the term in front of x.
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I have a 2.
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We are going to take that coefficient, divided by 2 and squared.
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It gets a lot of questions like, should you always divide by 2 and square it?
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The answer is yes it is always divided by 2 and squared.
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2 ÷ 2 is 1 and 1² is 1.
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This number right over here is what I need to add to both sides of my equation.
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There is all the original numbers and there is that +1 on both sides, just to keep things balanced.
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The reason why we are doing that is because on the left side it will now factor nicely
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You could use the reverse foil method on that left side and you will get x + 1 and x + 1.
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We just get a 2.
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If you do add the correct number to both sides, these factors will always be the same.
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If they are not the same then go back through and check your work.
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Make sure you did not make a mistake, but they will always end up being the same.
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We are going to write this as x + 1².
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That condenses things down and actually at this point this is where I can use my principle of square roots.
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We are going to use it because now I can take the square root of both sides to get rid of my square.
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The x + 1 = √2 , I will have x + 1= -√2.
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I can solve each of these directly.
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x = -1 + √2 and x= -1 -√2 .
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That one does take a lot more work, especially going through the entire process of completing the square.
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But notice how it did find our solutions.
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It even found our solutions, despite the fact that they were irrational solutions.
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Unfortunately completing the square can be a clunky process, and that is what I want to demonstrate with this next example.
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2x² – 1 = 3.
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Watch out even though the numbers are little bit more difficult to work with,
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I will still go through all of the same steps in order to get my solution.
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The very first thing I want to do is just get all of my x’s to one side.
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If it does not have an x, I will move it to the other.
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2x² - 3x we will add the 1 to the other side 1.
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Now I need to check my x² term to see if it is 1.
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It looks like this one is a 2 so I need to divide everything by 2.
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That will give us x² – 3/3x = ½.
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Notice how we now have a lot more fractions showing up.
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We have another method wrong it is just what the numbers turns out to be.
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Now that we have gotten to this point I need to figure out what number I should add to both sides of my equation
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so that the left side will factor nicely.
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It comes from this number right here.
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We will take it, divided by 2 and take that result and square it.
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Just because it is a fraction it does not mean we can not work with it,
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is the same as multiplying by ½ , so I get -3/4² which is the same as 9/16.
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9/16 is the number that I should add to both sides.
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Looking good, now that we have gotten to this stage the left side over here should factor nicely.
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You could use method of reverse foil to help you out.
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Know that your first terms will be x and two things that multiply to give you 9 would be a 3.
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And two things that will multiply to give you 16 would be 4.
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Both of these must be negative since our middle term is negative.
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Let us see what is going on the other side.
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I have ½ and I have 9/16.
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What I can get those together by finding a common denominator, 8/16 + 9/16.
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At this stage we have x -3/4 all of that is being squared and then all of that is equal to 17/16.
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Okay so completing a square is a lot of work for us.
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It gets our x’s together on the same side.
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We can solve directly using the principle of square roots and taking the square root of both sides.
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x - 3/4 I have the √17/16 or could be negative.
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I’m going to represent those by saying it could be the + or minus of that square root.
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I have one of my properties that says I can split up my square root over the top and bottom of a fraction.
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The √17 or √16 or x -3/4 = ±√17÷4.
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We are getting close to our actual solution, there is only one last thing I need to do.
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Let us go ahead and add 3/4 to both sides.
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I have x = ¾ ±√17÷4.
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Since these have the same denominator I will go ahead and just write them as one large fraction.
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That was quite a bit of work but notice how our answer was not a very nice answer.
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It happens to be another irrational number.
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We have no hope of finding irrational solutions like that using simply factoring.
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Completing the square is an important process for manipulating quadratics and still being able to find solutions.
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I will admit though it is quite a lengthy process, so be careful that you follow all the steps correctly.
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A much faster method than completing the square is our quadratic formula.
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The good news is, it will also work on any type of quadratic.
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You just have to be careful to pickup the proper values and put them in their spot.
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To help out, I’m going to write down the quadratic formula up in the corner here.
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x= - b ±√(b² )- 4ac ÷2a.
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That way we can reference it when we need to solve this one.
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In order for this to work out I need it set equal to 0 first.
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Let us get everything on one side.
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I’m going to organize things from the largest power to the smallest power.
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We will have to subtract that 7 over and now it is set equal to 0 just fine.
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My A value would be 3, B value would be 2 and the C value would be -7.
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I’m going to take all of those and put them into the formula.
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x is equal to negative the value of B which I said was 2 + or minus the square root.
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That is where we will put in B again 2² -4 × the value of A × the value of C.
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A and C all over 2 × the value of A.
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You will see I just substituted the proper values into all of their spots.
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We have to go through and very carefully simplify this down.
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We have -2 ± 2² = 4.
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I have that -4 × 3 × -7 I have to put all of those together.
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What will that give me? I have 84 ÷ 6.
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- 2 ±√88÷6.
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It is important to go ahead and try and simplify this as much as possible.
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I have √88 in there if you look at that as 4 × 22 and be able to take care of √4 and √2 separately.
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√4 would be 2 and √22 has to stay and then I can cancel out 2 from the top and the bottom.
00:24:41.300 --> 00:24:50.300
-1±√22÷3.
00:24:50.300 --> 00:24:55.500
You will notice how this answer was not so nice but the quadratic formula helped us find it.
00:24:55.500 --> 00:25:00.000
You just have to be very careful that you put it in the right spot and simplify carefully.
00:25:00.000 --> 00:25:09.200
One common mistake is watching your signs for the part underneath the square root.
00:25:09.200 --> 00:25:15.300
One last thing to do is to use our discriminate to determine the types of solutions we will get.
00:25:15.300 --> 00:25:18.000
No matter what method we use by the way.
00:25:18.000 --> 00:25:21.700
The discriminate is just that part underneath the square root.
00:25:21.700 --> 00:25:28.600
Think of b² – 4 × ac that is our discriminate.
00:25:28.600 --> 00:25:32.200
We can use this even though the numbers might get a little large.
00:25:32.200 --> 00:25:36.800
Here is my a, b, and c.
00:25:36.800 --> 00:25:41.600
Let us just evaluate what this discriminate would be.
00:25:41.600 --> 00:25:57.700
I have b², just 70, -4 × a × c.
00:25:57.700 --> 00:26:06.300
If I looked at 70², 7 × 7 is 49.
00:26:06.300 --> 00:26:10.800
I have some 0s in there and then 4 × 25 would be 100.
00:26:10.800 --> 00:26:15.800
100 × 49 = 49.
00:26:15.800 --> 00:26:21.700
I'm getting for this first discriminate is that it is all equal to 0.
00:26:21.700 --> 00:26:24.500
What is does that mean in terms of my solutions?
00:26:24.500 --> 00:26:32.200
If I was to look at the rest of the quadratic formula in there it is telling me that this entire part is going to 0.
00:26:32.200 --> 00:26:34.500
Either be adding 0 or subtracting 0.
00:26:34.500 --> 00:26:44.900
I will have one real solution.
00:26:44.900 --> 00:26:49.400
Let us try out the next one.
00:26:49.400 --> 00:27:03.100
I have a =1, b=4, c=2.
00:27:03.100 --> 00:27:12.800
B will be 4, -4a and c and they we will multiply all that together.
00:27:12.800 --> 00:27:24.800
4² is 16, 4 × 2 is 8, 16 - 8 is 8.
00:27:24.800 --> 00:27:39.400
This number is greater than 0 but it is not a square number.
00:27:39.400 --> 00:27:47.500
What I can say is that I will have two real solutions.
00:27:47.500 --> 00:27:56.900
Since it is not a square number they will be irrational.
00:27:56.900 --> 00:27:59.700
Let us do one more.
00:27:59.700 --> 00:28:11.800
Identify a, b, and c and we will put that into our discriminate.
00:28:11.800 --> 00:28:17.300
b² - 4 × a × c.
00:28:17.300 --> 00:28:27.500
It is time to crunch it all down, -10² is 100 and I have -4 × 3 × 15.
00:28:27.500 --> 00:28:42.300
I got 12 × 15, 2 × 1, 2 + 1.
00:28:42.300 --> 00:28:48.900
I have 100 - 180 or - 80.
00:28:48.900 --> 00:28:53.300
In this one our discriminant is less than 0.
00:28:53.300 --> 00:29:06.900
That is an indication that we will have two imaginary solutions.
00:29:06.900 --> 00:29:12.600
By using the discriminant you get a good idea of the solutions that you will get in your final answer.
00:29:12.600 --> 00:29:17.700
You can use a variety of techniques to go ahead and solve those quadratics and get your final answer.
00:29:17.700 --> 00:29:24.100
Know that factoring is usually the fastest, but quadratic formula will work on any type of quadratics.
00:29:24.100 --> 00:29:27.000
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