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Welcome to part two of[br]the presentation on

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quadratic equations.

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Well, I think I thoroughly[br]confused you the last time

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around, so let me see if I[br]can fix that a bit by doing

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several more examples.

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So let's just start with[br]a review of what the

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quadratic equation is.

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The quadratic equation says, if[br]I'm trying to solve the

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equation Ax squared plus Bx[br]plus C equals 0, then the

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solution or the solutions[br]because there's usually two

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times that it intersects the[br]x-axis, or two solutions for

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this equation is x equals minus[br]B plus or minus the square root

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of B squared minus[br]4 times A times C.

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And all of that over 2A.

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So let's do a problem and[br]hopefully this should make

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a little more sense.

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That's a 2 on the bottom.

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So let's say I had the equation[br]minus 9x squared minus

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9x plus 6 equals 0.

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So in this example what's A?

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Well, A is the coefficient[br]on the x squared term.

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The x squared term is here,[br]the coefficient is minus 9.

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So let's write that.

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A equals minus 9.

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What does B equal?

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B is the coefficient on the x[br]term, so that's this term here.

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So B is also equal to minus 9.

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And C is the constant term,[br]which in this example is 6.

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So C is equal to 6.

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Now we just substitute these[br]values into the actual

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quadratic equation.

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So negative B, so it's[br]negative times negative 9.

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That's B.

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Plus or minus the square root[br]of B squared, so that's 81.

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Right?

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Negative 9 squared.

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Minus 4 times negative 9.

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That's A.

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Times C, which is 6.

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And all of that over 2[br]times negative 9, which

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is minus 18, right?

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2 times negative 9-- 2A.

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Let's try to simplify[br]this up here.

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Well, negative negative[br]9, that's positive 9.

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

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Let's see.

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This is negative 4[br]times negative 9.

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Negative 4 times negative[br]9 is positive 36.

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And then positive 36[br]times 6 is-- let's see.

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30 times 6 is 180.

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And then 180 plus[br]another 36 is 216.

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Plus 216, is that right?

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180 plus 36 is 216.

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All of that over 2A.

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2A we already said is minus 19.

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So we simplify that more.

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That's 9 plus or minus the[br]square root 81 plus 216.

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That's 80 plus 217.

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That's 297.

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And all of that over minus 18.

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Now, this is actually-- the[br]hardest part with the quadratic

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equation is oftentimes just[br]simplifying this expression.

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We have to figure out if we[br]can simplify this radical.

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Well, let's see.

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One way to figure out if a[br]number is divisible by 9 is to

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actually add up the digits[br]and see if the digits

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are divisible by 9.

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In this case, it is.

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2 plus 9 plus 7 is equal to 18.

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So let's see how many[br]times 9 goes into that.

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I'll do it on the side here; I[br]don't want to be too messy.

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9 goes into 2 97.

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3 times 27.

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27-- it goes 33 times, right?

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

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of 9 times 33 over minus 18.

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And 9 is a perfect square.

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That's why I actually wanted to[br]see if 9 would work because

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that's the only way I could get[br]it out of the radical, if

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it's a perfect square.

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As you learned in that exponent[br]rules number one module.

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So this is equal to 9 plus[br]or minus 3 times the square

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root of 33, and all of[br]that over minus 18.

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We're almost done.

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We can actually simplify it[br]because 9, 3, and minus 18

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are all divisible by 3.

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Let's divide everything by 3.

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3 plus or minus the square[br]root of 33 over minus 6.

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And we are done.

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So as you see, the hardest[br]thing with the quadratic

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equation is often just[br]simplifying the expression.

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But what we've said, I know you[br]might have lost track-- we did

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all this math-- is we said,[br]this equation: minus 9x

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squared minus 9x plus 6.

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Now we found two x values that[br]would satisfy this equation

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and make it equal to 0.

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One x value is x equals[br]3 plus the square root

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of 33 over minus 6.

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And the second value is[br]3 minus the square root

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of 33 over minus 6.

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And you might want to[br]think about why we have

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that plus or minus.

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We have that plus or minus[br]because a square root could

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actually be a positive[br]or a negative number.

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Let's do another problem.

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Hopefully this one will[br]be a little bit simpler.

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Let's say I wanted to[br]solve minus 8x squared

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plus 5x plus 9.

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Now I'm going to assume that[br]you've memorized the quadratic

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equation because that's[br]something you should do.

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Or you should write it[br]down on a piece of paper.

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But the quadratic equation is[br]negative B-- So b is 5, right?

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We're trying to solve that[br]equal to 0, so negative B.

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So negative 5, plus or minus[br]the square root of B squared-

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that's 5 squared, 25.

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Minus 4 times A,[br]which is minus 8.

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Times C, which is 9.

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And all of that over 2 times A.

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Well, A is minus 8, so all[br]of that is over minus 16.

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So let's simplify this[br]expression up here.

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Well, that's equal to[br]minus 5 plus or minus

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the square root of 25.

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Let's see.

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4 times 8 is 32 and the[br]negatives cancel out, so

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that's positive 32 times 9.

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Positive 32 times 9, let's see.

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30 times 9 is 270.

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It's 288.

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I think.

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Right?

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288.

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We have all of that[br]over minus 16.

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Now simplify it more.

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Minus 5 plus or minus the[br]square root-- 25 plus

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288 is 313 I believe.

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And all of that over minus 16.

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And I think, I'm not 100% sure,[br]although I'm pretty sure.

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I haven't checked it.

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That 313 can't be factored[br]into a product of a perfect

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square and another number.

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In fact, it actually[br]might be a prime number.

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That's something that you[br]might want to check out.

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So if that is the case and[br]we've got it in completely

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simplified form, and we say[br]there are two solutions, two

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x values that will make[br]this equation true.

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One of them is x is equal[br]to minus 5 plus the square

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root of 313 over minus 16.

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And the other one is x is equal[br]to minus 5 minus the square

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root of 313 over minus 16.

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Hopefully those two examples[br]will give you a good

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sense of how to use the[br]quadratic equation.

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I might add some more modules.

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And then, once you master this,[br]I'll actually teach you how to

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solve quadratic equations when[br]you actually get a negative

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number under the radical.

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Very interesting.

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Anyway, I hope you can do the[br]module now and maybe I'll add a

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few more presentations because[br]this isn't the easiest module.

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But I hope you have fun.

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Bye.