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 go ahead and take a look at equations that are quadratic in form.
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This will be some special types of problems and that they are not quite quadratic,
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but we can manipulate them so that they at least look like quadratic.
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The reason why that is important is because then we can use our quadratic solving techniques.
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Let us go ahead and tackle them.
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We call that an equation is a quadratic equation if it is on the form ax² + bx + c =0.
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As long as a, that first value is not 0.
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If we can take an equation and write it in this form, then it is said to be in standard form.
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We will see many different examples of quadratic equations and how to solve them.
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There are a few different types of equations that are definitely not quadratic.
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For example, this one is x⁴ is not quadratic.
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This one has x²/3, the interesting thing about some of these equations is that you can treat them as quadratics during the solving process.
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The way you do this is you use a substitution process to go ahead and turn it into a much more manageable form.
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The way the substitution process works is that we choose part of the equation and we let that equal u.
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We will go ahead and we rewrite the entire equation using this new u.
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What you will find is that you should replace all of the original variables in the problem.
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If you had x’s now you just have a whole bunch of u’s.
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When you are done solving the equation though, we will make sure that we go back to the original variables.
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It is kind of tricky what I'm describing here, I want you to see it in action.
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You will see that it does make things much easier for solving these types of equations.
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Let us use one like this example.
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I want to solve x⁴ - 5x² + 4 = 0.
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This is not a quadratic because I’m dealing with x⁴ out front.
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What I'm going to do is I’m going to turn it into a quadratic form.
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I’m going to let u = x².
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The reason why I’m doing that is I'm hoping to replace this entire variable right there with u.
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If that was the only thing I replaced I would still be in trouble.
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I still have x's and u’s running around.
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Let us take that little equation that we created u = x².
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Let us square both sides of that.
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That would give us that u² = x⁴ and that gives us a way that we can also swap out this other x over here.
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x⁴ is the same as u².
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I can take our equation here and write it as u² - 5u + 4 = 0.
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The reason why that is just so important is because this new equation is quadratic in u.
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I have open many different solving techniques that I can use on this thing.
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I could try reverse foil to simply factor and use the 0 factor property.
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I could try our quadratic formula and get the solution directly, but I have lots of things open to me.
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This one is not too bad, so I'm just going to go ahead and factor it and then use that 0 factor property.
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My first terms must be a u and u.
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I need two things that multiply to get 4, but they add to give me -5.
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I only have one option for that, just -4 and -1.
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I know that u -4 could be 0 and u -1 = 0.
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Solving each of the separately would give me u = 4 and u = 1.
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Once we switch into a quadratic it is much more easier to manage.
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The problem is that we solve it for u in and our original problem had x’s.
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We are not exactly done with this problem yet.
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At this stage we want to borrow what we called u and swap out for our original x's.
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This says that x² = 4.
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This one is x² = 1.
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I'm left with two smaller quadratic equations and I will solve each of these directly.
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I will simply take the square root of both sides, so + - square root of 4 and x = + - the square root of 1.
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Good thing both of these can be simplified.
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+ -2 and + -1, so I have four different solutions for this particular problem.
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What got our foot in the door was being able to write all of these variables using a new variable and reducing those powers.
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You can do this for a variety of types of equations to put them in a quadratic form.
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What we want to recognize is what you should swap out for that u value.
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You can usually use this technique if you recognize that one of the powers is twice as large as the other power.
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2/3 is exactly twice as large as 1/3.
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I’m going to use this to help me swap out my x's for u’s.
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Let u =x¹/3.
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I remember that you can often figure out what the other one means to be by squaring both sides of this little equation.
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u² = (x¹/3)² which is the same as x²/3.
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I can swap out both of these.
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u² – 2u -15 = 0.
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Everything else is the same, but now I have those u’s in there and now it is quadratic in form.
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I can use a lot of other techniques to go ahead and solve this one.
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What shall we do with this one?
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I think this is another one we can simply factor without too much problem.
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My first terms better be u and now I need two things that will multiply and give me -15 but somehow add to get -2.
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Let us use 5 and 3.
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-5 and 3.
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This will give me that u -5 = 0, and u + 3 = 0.
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That comes from the 0 factor property.
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Solving each of the separately I have u = 5 and u = - 3.
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Each of these are looking pretty good.
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We want to go back to our original variable.
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I know exactly what u is, let us go ahead and put that in for both of these spots on here.
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x¹/3 x¹/3 and this is equal to 5 and equal to -3.
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To solve this directly from here, I think I can take both sides and cube it.
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5³ would be 125, -3³ = -27.
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Just like the foil, now we have our solutions to the original problem.
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We work all the way back to x, which was the original variable.
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Now some might actually be quadratic but you can still use this technique to put it in a much nicer form.
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In this next one, we are dealing with a 3x - 1² + 2 × 3x - 1 = 8.
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One way that you could handle this is simply to multiply everything out
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and then get everything on one side sand set it equal to 0 and use the quadratic formula.
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I'm not going to do that because I noticed that I actually have this common piece of 3x – 1.
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I'm going to swap out that common piece and call it something else.
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You will say let u = 3x - 1.
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Now that we have that, this will become much easier to solve.
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We will have a u² + 2 × u = 8.
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We can work to get everything on one side and now it is set equal to 0.
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For this u, let us go ahead and reverse foil it.
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u and u, need two numbers that would multiply to be -8 but add to be a positive 2, 4 and -2.
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This gives us two solutions u = -4 or u = 2.
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We must go back to that original variable.
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We know what u is, u = 3x – 1.
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We are going to put that in for both of our u’s.
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3x – 1 = -4 and 3x - 1 = 2.
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We can solve each of these separately.
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Adding 1 to both sides would be -3 and divide both sides by 3 would give us -1.
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There is one of our solutions right there.
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For the other one, we will add 1 to both sides and get 3, then divide both sides by 3 and get 1.
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We have our solutions for this problem.
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Let us do one more and this would involve some negative exponents.
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This one has 2 - y -6¹ - 1 = 6 × y – 6⁻².
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This one is a little jumbled up, it is difficult to figure out what we should swap out.
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It is good to know that we do have a y -6 that seems to be a common.
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That will be a part of what we end up exchanging with u.
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Let us go ahead and get everything onto one side.
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I’m going to move everything over to the right side 6y - 6⁻².
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I will add y -6⁻¹ -2.
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What I recognize here is that this -2 is exactly twice as large as that -1.
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It is a pretty good indication that I will end up swapping on my u for this piece right here.
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Let u = let us call this y – 6⁻¹.
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Let us go ahead and rewrite our equation using this.
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We have 6u² + u – 2.
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What I can see here is that it is definitely quadratic and the numbers are much smaller, much nicer to deal with.
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What do I do from here?
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I still have to be able to figure out what the solutions of this quadratic are.
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We got lots of tools available to us, let us go ahead and try the quadratic formula.
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-b ±√4, (±√b² – 4) × (a × c) ÷ 2a.
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I think this will simplify nicely.
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I have 4 × 6 = 24 × 2 = 48 ÷ 12 or -1 ± √49/12 or -1 ± 7/12.
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u = -1 + 7/12 and I have u = -1- 7/12.
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-1 + 7 = 6/12 would be -8/12.
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Both of those reduce, this one to ½ .
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The other one 4 goes into the top and 4 goes in the bottom, -3/2.
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I know little bit more about what u is equal to.
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Of course, we can not stop there, we must go back to our original variable.
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We must work back all the way to those y’s.
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Let us put those back in for u and see if we can solve this from here.
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Let us see y - 6⁻¹ = ½ and y - 6⁻¹ = -2/3.
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To solve this from here I think I will raise both sides to that -1 power.
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That will give me y - 6 and then with that -1 exponent, that will change the location of the 1 and 2, 2/1.
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Let us do the same thing with our other equation.
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Raise both sides to -1, this is y - 6 = -3/2, since it changes the location of the -2 and the 3.
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These are almost done.
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With the one on the left here, let us add 6 to both sides.
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This will give us y = 8 and with this other one let us add 6 to both sides, -3/2 + 6.
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It looks like we do have to get a common denominator, but this one is not so bad.
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It looks like it is just 9/2.
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We have worked all the way back to our original variable, we know what the solutions are.
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It can take quite a bit of work to end up swapping out the u and solving from there.
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The important part is that if we do not swap out the u these we do not have a lot of other techniques to solve equations like this.
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Make sure you properly identify what u you need to swap out and put it back in so you can get back to your original variable.
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