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Welcome to the presentation on[br]using the quadratic equation.

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So the quadratic equation,[br]it sounds like something

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very complicated.

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And when you actually first see[br]the quadratic equation, you'll

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say, well, not only does it[br]sound like something

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complicated, but it is[br]something complicated.

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But hopefully you'll see,[br]over the course of this

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presentation, that it's[br]actually not hard to use.

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And in a future presentation[br]I'll actually show you

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how it was derived.

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So, in general, you've already[br]learned how to factor a

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second degree equation.

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You've learned that if I[br]had, say, x squared minus

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x, minus 6, equals 0.

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If I had this equation. x[br]squared minus x minus x equals

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zero, that you could factor[br]that as x minus 3 and

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x plus 2 equals 0.

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Which either means that[br]x minus 3 equals 0 or

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x plus 2 equals 0.

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So x minus 3 equals 0[br]or x plus 2 equals 0.

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So, x equals 3 or negative 2.

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And, a graphical representation[br]of this would be, if I had the

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function f of x is equal to[br]x squared minus x minus 6.

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So this axis is[br]the f of x axis.

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You might be more familiar with[br]the y axis, and for the purpose

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of this type of problem,[br]it doesn't matter.

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And this is the x axis.

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And if I were to graph this[br]equation, x squared minus x,

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minus 6, it would look[br]something like this.

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A bit like -- this is f[br]of x equals minus 6.

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And the graph will kind of[br]do something like this.

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[br]34[br]00:01:57,15 --> 00:02:00,03[br]Go up, it will keep going[br]up in that direction.

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And know it goes through minus[br]6, because when x equals 0,

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f of x is equal to minus 6.

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So I know it goes[br]through this point.

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And I know that when f of x is[br]equal to 0, so f of x is equal

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to 0 along the x axis, right?

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Because this is 1.

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This is 0.

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This is negative 1.

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So this is where f of x[br]is equal to 0, along

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this x axis, right?

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And we know it equals 0 at the[br]points x is equal to 3 and

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x is equal to minus 2.

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That's actually what[br]we solved here.

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Maybe when we were doing the[br]factoring problems we didn't

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realize graphically[br]what we were doing.

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But if we said that f of x is[br]equal to this function, we're

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setting that equal to 0.

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So we're saying this[br]function, when does

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this function equal 0?

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When is it equal to 0?

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Well, it's equal to 0 at[br]these points, right?

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Because this is where[br]f of x is equal to 0.

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And then what we were doing[br]when we solved this by

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factoring is, we figured out,[br]the x values that made f of x

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equal to 0, which is[br]these two points.

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And, just a little terminology,[br]these are also called

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the zeroes, or the[br]roots, of f of x.

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[br]63[br]00:03:12,47 --> 00:03:14,81[br]Let's review that a little bit.

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So, if I had something like f[br]of x is equal to x squared plus

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4x plus 4, and I asked you,[br]where are the zeroes, or

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the roots, of f of x.

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That's the same thing as[br]saying, where does f of x

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interject intersect the x axis?

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And it intersects the[br]x axis when f of x is

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equal to 0, right?

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If you think about the[br]graph I had just drawn.

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So, let's say if f of x is[br]equal to 0, then we could

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just say, 0 is equal to x[br]squared plus 4x plus 4.

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And we know, we could just[br]factor that, that's x

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plus 2 times x plus 2.

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And we know that it's equal[br]to 0 at x equals minus 2.

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[br]78[br]00:04:10,17 --> 00:04:13,94[br]x equals minus 2.

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Well, that's a little[br]-- x equals minus 2.

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So now, we know how to find[br]the 0's when the the actual

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equation is easy to factor.

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But let's do a situation where[br]the equation is actually

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not so easy to factor.

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[br]85[br]00:04:32,12 --> 00:04:39,75[br]Let's say we had f of x[br]is equal to minus 10x

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squared minus 9x plus 1.

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Well, when I look at this, even[br]if I were to divide it by 10 I

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would get some fractions here.

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And it's very hard to imagine[br]factoring this quadratic.

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And that's what's actually[br]called a quadratic equation, or

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this second degree polynomial.

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But let's set it -- So we're[br]trying to solve this.

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Because we want to find[br]out when it equals 0.

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Minus 10x squared[br]minus 9x plus 1.

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We want to find out what[br]x values make this

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equation equal to zero.

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And here we can use a tool[br]called a quadratic equation.

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And now I'm going to give you[br]one of the few things in math

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that's probably a good[br]idea to memorize.

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The quadratic equation says[br]that the roots of a quadratic

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are equal to -- and let's say[br]that the quadratic equation is

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a x squared plus b[br]x plus c equals 0.

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So, in this example,[br]a is minus 10.

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b is minus 9, and c is 1.

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The formula is the roots x[br]equals negative b plus or minus

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the square root of b squared[br]minus 4 times a times c,

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all of that over 2a.

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I know that looks complicated,[br]but the more you use it, you'll

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see it's actually not that bad.

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And this is a good[br]idea to memorize.

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So let's apply the quadratic[br]equation to this equation

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that we just wrote down.

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So, I just said -- and look,[br]the a is just the coefficient

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on the x term, right?

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a is the coefficient on[br]the x squared term.

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b is the coefficient on the x[br]term, and c is the constant.

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So let's apply it[br]tot this equation.

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What's b?

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Well, b is negative 9.

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We could see here.

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b is negative 9, a[br]is negative 10.

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c is 1.

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Right?

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So if b is negative 9 -- so[br]let's say, that's negative 9.

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Plus or minus the square[br]root of negative 9 squared.

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Well, that's 81.

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[br]128[br]00:06:53,14 --> 00:06:56,94[br]Minus 4 times a.

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a is minus 10.

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Minus 10 times c, which is 1.

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I know this is messy,[br]but hopefully you're

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understanding it.

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And all of that over 2 times a.

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Well, a is minus 10, so[br]2 times a is minus 20.

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So let's simplify that.

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Negative times negative[br]9, that's positive 9.

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Plus or minus the[br]square root of 81.

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We have a negative 4[br]times a negative 10.

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This is a minus 10.

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I know it's very messy,[br]I really apologize

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for that, times 1.

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So negative 4 times negative[br]10 is 40, positive 40.

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Positive 40.

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And then we have all of[br]that over negative 20.

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Well, 81 plus 40 is 121.

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So this is 9 plus or[br]minus the square root

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of 121 over minus 20.

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Square root of 121 is 11.

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So I'll go here.

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Hopefully you won't lose[br]track of what I'm doing.

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So this is 9 plus or[br]minus 11, over minus 20.

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And so if we said 9 plus 11[br]over minus 20, that is 9

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plus 11 is 20, so this[br]is 20 over minus 20.

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Which equals negative 1.

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So that's one root.

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That's 9 plus -- because[br]this is plus or minus.

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And the other root would be 9[br]minus 11 over negative 20.

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Which equals minus[br]2 over minus 20.

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Which equals 1 over 10.

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So that's the other root.

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So if we were to graph this[br]equation, we would see that it

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actually intersects the x axis.

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Or f of x equals 0 at the[br]point x equals negative

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1 and x equals 1/10.

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I'm going to do a lot more[br]examples in part 2, because I

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think, if anything, I might[br]have just confused

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you with this one.

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So, I'll see you in the[br]part 2 of using the

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quadratic equation.

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[br]