WEBVTT mathematics/math-analysis/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about completing the square and the quadratic formula.
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In this lesson, we will be working just with quadratic polynomials--that is, polynomials that have degree 2.
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Quadratics are of the form ax² + bx + c, degree 2; a, b, and c are constant real numbers, and a is not equal to 0.
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Otherwise it wouldn't be a quadratic anymore, because we would have knocked out that x².
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In the previous lesson, we talked all about finding the roots of polynomials.
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Since quadratics appear often in nature, we have to find their roots a lot.
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However, also from the previous lesson, we saw that finding roots and factors...it is not always an easy business to factor a polynomial.
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So, wouldn't it be nice if there was an easier way to find the roots of a quadratic than having to figure out the exact factors and all that?
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And it turns out that there is; this lesson is going to explore that method, which will make it easier for us.
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First, let's remind ourselves of something we learned long ago in algebra: consider solving x² = 1.
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Now, our first automatic response would be to take the square root of both sides, and we might get x = 1.
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But we have to remember that that is only have of the answer.
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Hopefully we remember that, when we take the square root of both sides, we have to also introduce a plus and a negative version.
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Remember, since (-1)² = 1, and 1² = 1, we actually have two answers for this, x = 1 and x = -1.
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When you square a negative, it loses its negative-ness and becomes a positive number.
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So, if we wanted, we could express this as x = ± 1, using this symbol right here, which we call the "plus/minus symbol."
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The previous idea is given by this important rule--we have to always remember this any time we end up taking the square root.
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Otherwise we will introduce mistakes: whenever we take the square roots on both sides of an equation,
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when we are doing algebra, we put a plus/minus on one side.
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So, for example, if we have x² = 1, we take the square root of both sides; we get √(x²) = ±√1,
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at which point we get x = ± 1, which we can unfold into x = +1 and x = -1, our two answers.
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These things might get us thinking, though; it is easy to solve equations that are in this form, x² = k.
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We get x = ±√k; so that is pretty easy--if we could somehow get a quadratic to look like that, we would be doing pretty well.
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So, what if we had a polynomial like x² - 16 = 0?
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Well, then it is really easy: we just toss that 16 over; we get x² = 16.
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The square root of both sides: √(x²) = ±√16; so we take the square root of both of those; x = ±4.
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We managed to find the answer--nice--it worked really easily when we have this x² - k = 0.
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So, this method works great for x² - k = 0, because we just move it over and get x² = k,
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at which point we take the square root and introduce that plus/minus.
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But we couldn't find the roots of x² - 2x - 3 with it, because we can't just move it over.
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So, that is too bad...or could we?...maybe there is a way.
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Let's say a little bird points out to us that x² - 2x - 3 equals (x - 1)² - 4.
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Well, at this point, it is really easy to solve for the roots now.
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We set x² - 2x - 3 equal to 0, and then we use this piece of information right here.
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We know that we can swap this and this; we have this right here, so we swap it out,
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because we were told by that little bird (and we trust that little bird) that (x - 1)² - 4 is the exact same thing.
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So, at this point, we move our 4 over; we take the square root of both sides; √4 is 2, and we have ±,
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because we have to introduce that plus/minus when we take the square root.
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x - 1 = ±2; we move the 1 over, and we get x = 1 ± 2, which is equal to 3 and -1.
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We have gotten both of the answers for this quadratic.
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Great; what if there was some way that we could do this for any quadratic?
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If we could get this form of something squared minus k, it would be easy to find roots for any quadratic.
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So, let's try to do this on 4x² + 24x + 9 = 0.
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We will see if we can find a method for...we will call it completing the square,
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because we are going from this form where there is a bunch of stuff to this nice thing that is squared.
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So, we will call it completing the square, because once we have a square, minus just a constant,
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it is really, really easy to be able to solve for what the roots have to be.
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So, let's move the 9 over first; we will get 4x² + 24x = -9.
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Now, how can we get 4x² + 24x to become (_x + _)²?
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Well, there is that pesky 4, still in front of that x²; so let's just start by getting rid of that 4.
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We divide out the 4, and we have x² + 6x = -9/4.
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4x²/4 becomes just x²; 24x/4 becomes 6x (6 times 4 is 24); that equals -9/4; great.
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From this format, we want to get some (x + _)², (x + r)².
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Now, notice: (x + r)² is equal to x² + 2rx + r²...
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sorry, (x + r)² becomes xr + rx; so we have two r's showing up on that x, 2r times x.
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x² + 2rx + r²: we need to figure out what goes in these blanks.
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We have a blank here; x² + 6x + effectively a blank...to be able to complete and get this here.
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So, (x + _)²--how can we do this? Well, we will use this information that we just had here.
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We realize that x² + 6x + 9 = (x + 3)².
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We notice that if it is going to be a blank here, and it has to connect to here, well, that was 2r here; so it must be just 1r here.
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So, if 2r is 6, then 1r is 3; and we check this out: x² + 6x + 9 = (x + 3)².
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We check it: x times x is x²; x times 3 is 3x, plus 3 times x is another 3x, so 6x; great; plus 3²...3 times 3 is 9; great.
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It checks out; so we have x² + 6x = -9/4; that is what our equation was.
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How do we get a 9 on the left? Simple--we just add a 9 to both sides.
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So, add 9 to both sides, because we figured out that we want it to look like this.
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Since we want it to look like this, we make it look like this through basic algebra manipulation.
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Add 9 to both sides; we get x² + 6x + 9 = -9/4 + 9.
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The left side is now (x + 3)²; we collapse it, and we have it equal to (-9 + 36)/4,
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since 9 is equal to 9 times 4, over 4, which equals 36/4.
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It connects to that other -9/4 by getting a common denominator and then adding to it.
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So, it is (-9 + 36)/4; we simplify that, and we get (x + 3)² = 27/4; great.
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If we wanted to, we could easily solve this.
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So, we take the square root of both sides: we get ±√27/√4 = x + 3--easy.
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We call this procedure, once again, completing the square, because we are going from a method
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that doesn't really have this nice squared chunk to a thing that does have this nice squared chunk,
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just minus some other factor or plus some other factor.
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We can do this in general; we can do this to some general quadratic polynomial, ax² + bx + c = 0.
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We can do this in general, and basically follow the exact same method that we just did with numbers.
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So, first we move the c over; just like we move the 9 over, we have -c now.
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Since we eventually want something of the form (x + _)², we don't want this pesky a getting in the way.
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So, we divide both sides by a; b divided by a becomes over a; divided by a becomes just a 1 in front of that x²;
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divide by a over here...-c/a; great, so we get x² + b/a(x) = -c/a.
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All right, next our goal is something of the form (x + _)².
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Now, we notice, once again: (x + r)², whatever r is, equals x² + 2rx + r².
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Now, we want to match up to this format; we already have b/a(x), and we want 2r(x).
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The x² here matches with the x² here; the 2rx here matches with the b/a(x) here;
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and the r² that we haven't introduced yet
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is what the blank is that we don't know what we are going to put in yet.
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So, if 2r is the same thing as b/a, if b/a = 2r, then that means b/2a = r.
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So now, with that in mind, we know that what we want to introduce is r²:
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b/2a = r, so we want to add r², or (b/2a)², to both sides.
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So, we add r² = b²/4a², and we complete the square.
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x² + b/a(x) + b²/4a²...that collapses into (x + b/2a)².
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Check that out really quickly: x times x becomes x²; great; x times b/2a becomes b/2a(x),
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plus it will happen a second time; so b/2a + b/2a becomes 2b/2a, so just b/a, still times x;
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b/2a times b/2a becomes b², over...2 times 2 is 4...a times a is squared; so it is b²/4a².
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Great; that checks out; and we added this b²/4a² to both sides; we can't just add it to one side.
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And so, that will collapse into (b² - 4ac)/4a², because we have -c/a; so that becomes -4ac/4a².
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So, we can get them on a common denominator; so we have completed the square.
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Great; that is a general form.
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At this point, we have shown that any equation that starts as ax² + bx + c = 0 is equivalent,
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through completing the square, to this equation right here.
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It is a little bit complex, but we just proved that we can just do that through basic algebraic manipulation.
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At this point, it would be quite easy to solve a given quadratic for x by plugging in values for a, b, and c, then just doing a little algebra.
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For example, if we have 4x² + 8x + 2, then our a is equal to 4; our b equals 8; and our c equals 2.
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Oh, let's color-code that; so a at 4 is red; b at 8 is blue; and green is c = 2; lovely.
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This is x + b over 2a; so our a's are the red things; our blues are the b's; and our c is that green.
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So, we follow this format; and we have blue 8² here; our a's...4 here and 4 here; 4 here;
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and then finally, our green is here, and this coefficient here just stays here; this coefficient here just stays here;
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this coefficient here just stays here; so if we wanted to, at this point, we could just do some arithmetic;
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and we would be able to simplify that, and then we would be able to take the square root,
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and basically just be able to solve for it, and we would be able to get the answer.
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But we can go one step farther, and we can just set up a general formula to solve any quadratic, ax² + bx + c = 0.
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We are so close to this; and then we can just use that formula in the future, any time we want to find the roots of any quadratic.
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So, at this point, we have shown that ax² + bx + c = 0 is the same thing;
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it is equivalent to the completed-square version of (x + b/2a)² = (b² - 4ac)/4a².
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So, we just take the square root of both sides to get to the x: x + b/2a = ±√(b² - 4ac...
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what is the square root of 4a²? 4 comes out as 2; a² comes out as a.
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So, we get the 2a on the bottom; so (x + b/2a) = ±√(b² - 4ac), all over 2a.
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Next, we isolate for x; we subtract the b/2a, plus or minus √(b² - 4ac), all over 2a.
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Look, they are already in common denominators; so we get x = [-b ± √(b² - 4ac)]/2a.
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We have the **quadratic formula**, an easy way to solve for the roots of any quadratic polynomial.
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So, as long as we have some quadratic polynomial like this, we just plug into this thing and do some arithmetic.
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It might get ugly; it might require a calculator; it might not be really easy arithmetic.
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But there is not much thinking that we have to do; there is no difficult cleverness of figuring out just the right way to factor it.
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We just plug in and go, and an answer will pop out.
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Now, I am not a big fan of memorizing a lot of things; I think, for the most part, that you want to understand how to get to these things.
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But the quadratic formula is going to come up so often that you are going to end up needing to see
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this [-b ± √(b² - 4ac)]/2a...I am going to have to recommend that you probably want to memorize this thing.
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Memorize the quadratic formula, because it will show up a lot.
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And even if your teacher doesn't absolutely require you to have it memorized in another class,
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you are going to end up seeing this so often, and you are going to have to solve for so many quadratics,
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that you want to just have this ready, so that you can pull it out any time.
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You will just remember and think, "Oh, yes, I am trying to look for the roots of a quadratic; I can solve this through the quadratic formula."
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It comes up a lot, so it is good to just have it memorized.
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All right, follow the format if you are going to use the formula.
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It is really important to note that, if you want to use the quadratic formula,
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the polynomial must be set up in this format, ax² + bx + c = 0.
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It absolutely has to be set up in this format.
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For example, if we have 2x² - 47x + 23, then our a equals 2; our b equals -47
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(because notice that here it is a +, but here it is a negative, so it must be a part of the number); and then finally, our c equals 23.
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So, a = 2; b = -47; c = 23; great.
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But it would be totally wrong, absolutely wrong, to say x² + 3x - 4 = -2x + 8 gives us a = 1, b = 3, c = -4.
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We have this equals...stuff over here; it has to equal 0; otherwise it doesn't work.
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It has to be in this format of ax² + bx + c = 0.
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That is how we derive completing the square; that is how we derive the quadratic formula.
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If it is not in this format, it breaks down entirely; we can't use the quadratic formula.
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So, we have to put it into this format before we can use the quadratic formula.
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It absolutely has to be in the form; otherwise, it just doesn't work.
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How many roots does a quadratic have?
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Now, of course, not all quadratics have the same number of roots.
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The graphs below show the three possibilities: we have one where it intersects it twice (one here, one here--two roots);
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we have it where it intersects it just once (it barely grazes and touches, barely just hitting it once);
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or we have absolutely none, where it never manages to cross the x-axis.
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And of course, these could all flip the other way; we could have it going down this way.
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We could have it barely touching on this side; and we could also have it crossing over like this.
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There is no guarantee that it has to be pointed or look like these.
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But the idea is just that these are the numbers of times it could cut: it could cut on both sides, cut just once at the tip,
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or cut not at all, because it never manages to cross it.
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So, the quadratic formula actually manages to show us which one of these situations we are in.
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The way we do this is through the discriminant.
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Remember the formula: -b ± √(b² - 4ac), all over 2a.
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You are going to hear that a lot; it is good to memorize.
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The expression b² - 4ac is the discriminant; it tells us how many roots there are.
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b² - 4ac being greater than 0 means that there are two roots.
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If it is equal to 0, then we have one root; and if it is less than 0, we have 0 roots,
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each corresponding to those colors on the last picture, as well.
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So, the discriminant tells us how many roots--what kind of situation we are in
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for how often our parabola is going to actually manage to cross over that x-intercept and touch that x-axis.
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Why or how does it work--what is going on here?
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Let's look at the quadratic formula once again.
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Remember: our discriminant is the b² - 4ac part, the part underneath the square root.
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It is under the square root; each of the three cases we just saw correlates to how many answers come out of the square root.
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We have this plus or minus here; so if b² - 4ac is positive, two are going to come out, because of the plus or minus.
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If we have the square root of 4 in there, then we have plus 2 and minus 2 (the square root of 4 is 2, so we have ± 2).
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So, we have a plus version and a minus version; that is two different worlds, so we get two different answers.
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But if we have b² - 4ac equals 0, then ± 0 is just 0.
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So either way, it is just the same world; √0...plus or minus 0...it doesn't really matter if we go with the plus or the minus; we get the same thing.
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So, we just have one root--only one possibility.
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Finally, if b² - 4ac is negative (that is, less than 0), the square root fails entirely, so there are no answers.
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It is impossible to take the square root of a negative number, remember, because any positive squared becomes positive;
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any negative squared becomes negative; so there is no number out there that, when you square it, will become a negative.
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So, there are no answers if b² - 4ac is negative--at least, there are no real answers.
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We will talk about how things get a little shady once we get into complex numbers.
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But for right now, the discriminant tells us that there are no answers if b² - 4ac is negative.
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Really, we can just look at how this interacts with square root--how many things can come out of ± square root.
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If two things can come out, because it is a positive number under the square root, then we have two possibilities, two answers.
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If 0 is under the square root, then there is one possibility, because it is just one possibility underneath √0.
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So, ±0 is just one thing; if it is ± the square root of a negative number,
00:18:20.600 --> 00:18:25.600
then that is impossible to do in the first place, so we have no possibilities under it; we have no answers.
00:18:25.600 --> 00:18:27.000
All right, we are ready for some examples.
00:18:27.000 --> 00:18:31.600
Complete the square on the polynomial 3x² - 30x + 87 to give an equivalent expression.
00:18:31.600 --> 00:18:35.200
Now, we didn't formally talk about completing the square when it wasn't equal to 0.
00:18:35.200 --> 00:18:45.100
But we can follow the exact same method: 3x² - 30x + 87--the first thing we do is...we want to look at that 87 as being off on the side.
00:18:45.100 --> 00:18:48.600
It is still part of the expression, but it is not what we really want to work with, fundamentally.
00:18:48.600 --> 00:19:00.100
Now, we have a 3 in here, so we want to pull this 3; we will pull it out; we get 3(x² -...30...pulling out a 3 gets 10x) + 87 (is still over there).
00:19:00.100 --> 00:19:05.500
3(x² - 10x); we check--yes, it checks out; we still have the same thing there.
00:19:05.500 --> 00:19:09.800
The next step: we want to figure out + _...what blanks can go in there?
00:19:09.800 --> 00:19:17.900
Well, remember: if it was (x + r)², then that would become x² + 2rx + r².
00:19:17.900 --> 00:19:22.600
So, what makes up our 2r right now? Our 2r here is our -10.
00:19:22.600 --> 00:19:31.100
So, if 2r = -10, then our r by itself would be equal to -5; so we want to get a 25 inside.
00:19:31.100 --> 00:19:44.500
So, we have 3(x² - 10x +...[we want r² equal to]25); -5 times -5 is 25, so it is +25 on the inside.
00:19:44.500 --> 00:19:48.800
So, we want that to show up; of course, if we just change our expression around--we just put a number in there--
00:19:48.800 --> 00:19:52.000
it is not the same expression; we just broke our mathematics--it doesn't work like that.
00:19:52.000 --> 00:19:56.700
So, we have to make sure that however much we put in on one place, we take out of somewhere else.
00:19:56.700 --> 00:20:02.400
I can have any number...5...and I could add 3 and subtract 3, and it would have no effect.
00:20:02.400 --> 00:20:05.300
I would still be left with 5, because the same thing is adding 0.
00:20:05.300 --> 00:20:11.400
So, if we add 25 into the inside of our quantity, how much do we need to take away on the outside?
00:20:11.400 --> 00:20:18.500
Well, putting in 25 on the inside...remember, it is 3 times (....+ 25).
00:20:18.500 --> 00:20:33.100
Well, that is going to be ..... + 75; 3 times 25 goes to connect like that; 2 times 25 means we need to take away 75
00:20:33.100 --> 00:20:39.800
on the outside to keep our scale balanced; we put 25 into the inside of the parentheses;
00:20:39.800 --> 00:20:43.500
3 times 25 is 75; so we need to take 75 out.
00:20:43.500 --> 00:20:49.200
We put a total of 75 in the expression; so if we take a total of 75 out, our scale remains balanced.
00:20:49.200 --> 00:20:55.400
- 75...and then we still have to bring on what was in the expression before, + 87.
00:20:55.400 --> 00:21:04.000
So, 3(x² - 10x + 25)...(x + r)²...our r equals -5, so we have (x - 5)².
00:21:04.000 --> 00:21:08.000
Let's check and make sure that is still correct: x², x times x, x².
00:21:08.000 --> 00:21:17.100
x times -5 is -5x; -5 times x is -5x again, so double that: -10x...it checks out still; -5 times -5 is 25; great, it checks out.
00:21:17.100 --> 00:21:27.400
Minus 75 + 87; that becomes + 12; and now we have something that is equivalent.
00:21:27.400 --> 00:21:33.000
3(x - 5)² + 12...let's check to make sure that is correct.
00:21:33.000 --> 00:21:50.900
3(x - 5)² would become x² - 10x + 25, plus 12; 3x² - 30x + 75 + 12; 3x² - 30x + 87.
00:21:50.900 --> 00:21:53.800
Great--it checks out; that is the same thing as what we started with.
00:21:53.800 --> 00:22:00.600
All right, the next one--the second example: Solve -x² + 10x - 20 = 4x - 16.
00:22:00.600 --> 00:22:04.000
So, we have that nice, fancy quadratic formula; let's try it out.
00:22:04.000 --> 00:22:08.700
The first thing, though: it is not currently equal to 0--it is not set to 0,
00:22:08.700 --> 00:22:13.600
so we need to get the whole thing so it looks like that ax² + bx + c = 0.
00:22:13.600 --> 00:22:27.700
So, let's move things around: we subtract 4x; we add 16; we get -x² + 6x - 4 = 0.
00:22:27.700 --> 00:22:41.100
Great; so we are now set up--we have a = -1, b = 6, c = -4; we are in that format.
00:22:41.100 --> 00:22:47.000
Our normal format is ax² + bx + c = 0; so we have that parallel.
00:22:47.000 --> 00:23:02.200
What is our formula? The roots are going to be x = [-b ± √(b² - 4ac)]/2a.
00:23:02.200 --> 00:23:17.000
All right, so we start plugging into that: we have x =...plug in our blue -6, our b; plus or minus the square root...
00:23:17.000 --> 00:23:30.000
b² is 6²; minus 4 times a (-1) times c (-4)...keep that going;
00:23:30.000 --> 00:23:36.700
the colors are getting a little bit crazy here, but the basic idea going on is still the same;
00:23:36.700 --> 00:23:49.500
-6 ± √...2 times...and the red one here, a; and notice these coefficients--they just stick around the whole time; -4, -4.
00:23:49.500 --> 00:23:56.500
They just stick around no matter what; the ± moves down; the negative moves down; so those things are always there.
00:23:56.500 --> 00:24:07.200
At this point, all we have to do is solve it out; so x = [-6 ± √(36...(-4)(-1)(-4)...
00:24:07.200 --> 00:24:15.600
two of those cancel out, but we are still left with a negative, so...- 16, over 2 times...
00:24:15.600 --> 00:24:21.300
we are going to replace that with what it should become, 2 times -1; I got confused by all the colors.
00:24:21.300 --> 00:24:34.700
2 times -1 becomes -2; so we have x = -6/-2 ± √(20)/-2.
00:24:34.700 --> 00:24:51.000
x =...this becomes positive 3, plus or minus the square root of 20 (is equal to √(4)(5)); we get 2√5.
00:24:51.000 --> 00:24:58.900
So, we replace that down here; so we have 3 ± 2√5, all over -2.
00:24:58.900 --> 00:25:05.500
That is 3; now, the negative here hits that plus/minus, but all that is going to do is cause the plus to become a negative,
00:25:05.500 --> 00:25:10.600
and the negative to become a plus, so it didn't really do anything: plus/minus is the same thing as minus/plus.
00:25:10.600 --> 00:25:13.100
We are basically where we were before: √5.
00:25:13.100 --> 00:25:22.400
So, our answers are going to be x = 3 - √5 and 3 + √5; those are the solutions to that,
00:25:22.400 --> 00:25:29.800
because we found the roots to when we turn it into the format that we could use it on; great.
00:25:29.800 --> 00:25:36.000
A man standing on the top of a 127-meter-tall cliff throws a ball directly down at 10 meters per second.
00:25:36.000 --> 00:25:45.700
The height of the ball above the ground is given by: height at time t equals -4.9t² - 10t + 127, where t is in seconds.
00:25:45.700 --> 00:25:48.500
How long does it take for the ball to hit the ground?
00:25:48.500 --> 00:25:55.500
We have this man; he is standing on top of a cliff; and for some reason, he throws a ball down.
00:25:55.500 --> 00:26:01.300
All right, the ball is going down; it is moving down towards the ground down here.
00:26:01.300 --> 00:26:07.200
Let's see what this means: what is ground?
00:26:07.200 --> 00:26:10.200
Ground is h =...what does that mean?
00:26:10.200 --> 00:26:14.200
Well, we notice that if we plug in 0, then that is going to be just as he threw it, which would be...
00:26:14.200 --> 00:26:16.400
the 0's would cancel out; the t would cancel with the t here;
00:26:16.400 --> 00:26:22.800
and we would be left with just 127, which is where he starts at, 127 meters high.
00:26:22.800 --> 00:26:27.500
So, that makes sense--that the ground is going to be 0 meters high; it makes intuitive sense.
00:26:27.500 --> 00:26:31.600
So, we know that what we are looking for is when the ball hits the ground.
00:26:31.600 --> 00:26:38.800
When does the ball hit the ground? The ball hits the ground at h = 0.
00:26:38.800 --> 00:26:43.500
That is what the height we are looking for is: ground is normally put at h = 0.
00:26:43.500 --> 00:26:45.800
You can sometimes move it around; but for the most part, you will end up seeing,
00:26:45.800 --> 00:26:48.400
in any word problem where they are talking about the height, that the ground level--
00:26:48.400 --> 00:26:53.200
and normally, whatever our base level is considered--is a height of 0.
00:26:53.200 --> 00:26:55.100
How long does it take for the ball to hit the ground?
00:26:55.100 --> 00:27:00.600
Well, that means, if we are looking for when h equals 0, that we are going to use that in here in our functions.
00:27:00.600 --> 00:27:10.800
So, 0 = -4.9t² - 10t + 127; we plug this into our quadratic formula.
00:27:10.800 --> 00:27:14.600
We are not looking at x anymore; we are looking at when t is going to give our roots.
00:27:14.600 --> 00:27:39.500
So, t = -b, -(-10), plus or minus the square root of (-10)², minus 4ac, 4 times -4.9 times 127, all over 2a, 2 times -4.9.
00:27:39.500 --> 00:27:54.400
t = +10 ± √(100 -...that will end up...we work that out with a calculator...2489.2), over -9.8.
00:27:54.400 --> 00:27:59.300
So, at this point, that is not super easy to work out; so we are going to start plugging into a calculator.
00:27:59.300 --> 00:28:03.200
We won't actually do that here; but we are going to basically plug two different expressions into our calculator.
00:28:03.200 --> 00:28:17.000
The plus version is...one mistake there...that was -4 times -4.9, so it becomes +.
00:28:17.000 --> 00:28:20.000
Otherwise it would be impossible, because we would have a negative underneath our square root.
00:28:20.000 --> 00:28:22.400
That is what made me catch that.
00:28:22.400 --> 00:28:37.000
2589.2 divided by -9.8...and 10 plus 10 minus √2589.2/-9.8.
00:28:37.000 --> 00:28:48.100
We plug that into our calculator, and we get two different answers: t = -6.213 and 4.172.
00:28:48.100 --> 00:28:51.300
Now, at first, that should set off some alarm bells in our head.
00:28:51.300 --> 00:28:55.700
We throw a ball down a cliff...we imagine this in our head; that is the very first step--we imagine it in our head.
00:28:55.700 --> 00:28:59.400
You throw a ball down some tall height, and eventually it hits the ground.
00:28:59.400 --> 00:29:03.000
All right, that is the end; it doesn't hit the ground at two different times.
00:29:03.000 --> 00:29:11.500
And so, what does this mean?--we have two different answers here, so which one is the correct answer, -6.213 or 4.172?
00:29:11.500 --> 00:29:17.900
Which one is right? We think; we know that this is t given in seconds, so we know what happens.
00:29:17.900 --> 00:29:23.800
He throws the ball down, and so forward in time, it is falling; we are going by this equation.
00:29:23.800 --> 00:29:28.600
But what about a negative time--did he throw the ball before 0 seconds?
00:29:28.600 --> 00:29:33.000
No, he throws the ball at 0 seconds; it starts at his height at 0 seconds.
00:29:33.000 --> 00:29:41.300
So, it must be that h(t) is only true--its domain is only 0 to positive infinity.
00:29:41.300 --> 00:29:48.200
And actually, it is not even going to be true after 4.172, because all of a sudden the ground gets in the way and stops this equation from being true.
00:29:48.200 --> 00:29:58.800
So, h(t) is only true from t = 0 until the ball hits the ground--until whatever t the ball hits the ground at.
00:29:58.800 --> 00:30:04.300
So, that is sort of an implicit thing that we hadn't explicitly stated; but we had to understand what is going on,
00:30:04.300 --> 00:30:07.200
because otherwise we will get two answers, and one of them is going to be wrong.
00:30:07.200 --> 00:30:12.100
We have to realize what is going on; we don't want to just blindly do what the formula told us.
00:30:12.100 --> 00:30:16.000
We want to think about what this represents; word problems require thinking.
00:30:16.000 --> 00:30:23.200
So, -6.213 and 4.172...we realize it is only true from t = 0 to higher numbers, until the ball hits the ground.
00:30:23.200 --> 00:30:30.800
The ball hits the ground at 4.172; and -6.213 is an extraneous solution--it is impossible to look at those times,
00:30:30.800 --> 00:30:37.500
because that is back before this function ever ended up even being used.
00:30:37.500 --> 00:30:42.500
The function comes into existence only once the ball is thrown at time t = 0.
00:30:42.500 --> 00:30:45.900
So, negative times are completely extraneous--we can't use those answers.
00:30:45.900 --> 00:30:53.100
And we get 4.172 seconds as the correct answer; great.
00:30:53.100 --> 00:30:57.600
The final example: Two cars are approaching a right-angle intersection on straight roads.
00:30:57.600 --> 00:31:01.000
The first one is coming from the north at a constant speed of 30 meters per second,
00:31:01.000 --> 00:31:05.200
while the second one is from the east at a constant speed of 25 meters per second.
00:31:05.200 --> 00:31:10.800
If both cars are currently 200 meters from the intersection, how much time is there until they have a distance of 90 meters between them?
00:31:10.800 --> 00:31:15.900
This is the classic nightmare word problem with so many things here--what are we going to do?
00:31:15.900 --> 00:31:20.000
Well, we just start figuring out what it is telling us; then we will work on the math.
00:31:20.000 --> 00:31:24.500
So, the first thing we do is try to make a picture of a right-angle intersection.
00:31:24.500 --> 00:31:27.800
We know what an intersection looks like on the street.
00:31:27.800 --> 00:31:34.400
Two streets intersect one another; we know that they came together at a right angle, because it says "right-angle intersection."
00:31:34.400 --> 00:31:39.300
Great; they are on straight roads, so we are guaranteed the fact that straight lines come out of it; so this makes sense.
00:31:39.300 --> 00:31:44.400
The first one is coming from the north; well, let's put north as being this way.
00:31:44.400 --> 00:31:49.500
This car is up here; it is coming from the north at a constant speed of 30 meters per second.
00:31:49.500 --> 00:31:53.600
It is going down 30 meters per second; where is it right now?
00:31:53.600 --> 00:31:59.300
We know it is 200 meters away at the start; what about the other one?
00:31:59.300 --> 00:32:08.200
The other one is coming from the east (east will be over here), and it is going at 25 meters per second.
00:32:08.200 --> 00:32:17.200
How far away is it? We figured out the 200 meters to get the first; both cars are currently 200 meters, so it is 200 here, as well.
00:32:17.200 --> 00:32:20.100
How much time is there until they have a distance of 90 meters between them?
00:32:20.100 --> 00:32:26.900
We also want to be able to introduce...the other thing that we have to figure out is what it means by "distance between them."
00:32:26.900 --> 00:32:35.000
If we have a point here and a point here, then the distance is just the distance between those two points.
00:32:35.000 --> 00:32:39.200
But we don't have any information about the distance between them; we are not told how far away they are.
00:32:39.200 --> 00:32:44.500
But we are told how far they are from this intersection in the middle.
00:32:44.500 --> 00:32:53.600
If we have...here is x and here is y...look, we can use the Pythagorean theorem: x² + y² = d².
00:32:53.600 --> 00:32:55.700
Great; so we have a way of relating these two things.
00:32:55.700 --> 00:33:00.300
But if we are just frozen at 200, 200² + 200² = distance squared,
00:33:00.300 --> 00:33:04.800
then we are going to get something that is never going to be 90, because we are just looking at a single snapshot in time.
00:33:04.800 --> 00:33:08.600
We also have to have a way of their movement, their motion, affecting this.
00:33:08.600 --> 00:33:13.500
So, they start at 200 away; but then they start to get closer and closer to that intersection.
00:33:13.500 --> 00:33:20.000
It is going to be 200 - 30t, because it is going to be that, as they get closer to the intersection,
00:33:20.000 --> 00:33:24.300
as more time goes on, more of their distance from the intersection will disappear.
00:33:24.300 --> 00:33:34.500
So, the 200 - 30t...and then, the other one is 200 - 25t, because the red one, the north car,
00:33:34.500 --> 00:33:39.600
is coming at 30 meters per second; so for every second that goes by, it will have moved 30 meters
00:33:39.600 --> 00:33:46.400
towards the intersection, so it will be 200 - 30t; and the blue car, the east car, will be 200 - 25t,
00:33:46.400 --> 00:33:53.700
because for every second, it moves 25 meters towards the intersection: 200 less 25 meters.
00:33:53.700 --> 00:33:56.500
OK, so at this point, we have a real understanding of what is going on.
00:33:56.500 --> 00:34:03.500
We can now put this d in here; and we can connect all of these ideas.
00:34:03.500 --> 00:34:09.400
So, we have 200 - 30t is what describes the red car, our north car.
00:34:09.400 --> 00:34:15.000
And 200 - 25t is what describes the east car, our blue car.
00:34:15.000 --> 00:34:19.700
And we are looking for when the distance is equal to 90.
00:34:19.700 --> 00:34:28.100
So remember: we had x² + y² = d² from the fact that this is just a nice, normal Pythagorean triangle.
00:34:28.100 --> 00:34:35.300
We can use the Pythagorean theorem here: the square of the two sides is equal to the square of the hypotenuse.
00:34:35.300 --> 00:34:46.000
So, we have 90² = (200 - 30t)² (our red car--our north car) + (200 - 25t)² (our east car--our blue car); great.
00:34:46.000 --> 00:34:49.200
And now we are trying to solve this and figure out when t is going to make this true.
00:34:49.200 --> 00:34:52.500
At what t will that equation there be true?
00:34:52.500 --> 00:34:56.700
So, we just start working it out; now, this is going to get pretty big pretty fast, because we have big numbers.
00:34:56.700 --> 00:34:59.400
But luckily, we have access to calculators in this world.
00:34:59.400 --> 00:35:09.800
So, 90²...plug that in; that comes out as 8100; 200 times 200 becomes 40000; minus 30t...
00:35:09.800 --> 00:35:16.300
200 times -30t, plus -30t times 200; 200 times -30t is -6000, but we have to double that,
00:35:16.300 --> 00:35:26.100
so it is -12000t; minus 30t times -30t...+ 900 t².
00:35:26.100 --> 00:35:35.400
The next one is + 40000 again for the other portion; 200 times -25 becomes -5000.
00:35:35.400 --> 00:35:43.700
But then, we also have to have -25 times 200 again the other way, so it is -500 doubled, so -10000t.
00:35:43.700 --> 00:35:52.200
Plus -25 times -25, so + 625t²...
00:35:52.200 --> 00:35:57.800
All right, we simplify this; this looks like something that could eventually turn into a quadratic.
00:35:57.800 --> 00:36:00.100
So, we say, "Oh, it is time to use the quadratic formula!"
00:36:00.100 --> 00:36:10.400
So, we need to get into that form: we will subtract 8100 over, so -8100, -8100 over here...
00:36:10.400 --> 00:36:18.900
We look at 625t² + 900t²; we get 1525t².
00:36:18.900 --> 00:36:33.000
Next, -10000t; -12000t; -22000t; 40000 and 40000...minus 81000, plus 71900; wow.
00:36:33.000 --> 00:36:35.500
But we are in a position where we can now use the quadratic formula.
00:36:35.500 --> 00:36:38.900
Once again, it is a good thing that we have calculators; otherwise this would be really difficult.
00:36:38.900 --> 00:36:41.000
But we can use the quadratic formula now.
00:36:41.000 --> 00:36:53.500
So, t = -b, -(-22000), so 22000, plus or minus the square root of b², 22000²;
00:36:53.500 --> 00:37:05.500
we will drop the negative sign, just because it is going to get squared anyway; minus 4, times 1525, times 71900.
00:37:05.500 --> 00:37:09.700
At this step, we might toss those parts in, just the part underneath the square root, use the discriminant,
00:37:09.700 --> 00:37:13.800
and make sure that there is an answer; it will turn out that there is an answer, so we will just keep going.
00:37:13.800 --> 00:37:19.200
And that is going to be divided by 2 times 1525.
00:37:19.200 --> 00:37:22.900
I won't work this all out here; I am going to trust the fact that you can do the two different versions.
00:37:22.900 --> 00:37:27.000
Remember: there is a plus version, and then there is a minus version.
00:37:27.000 --> 00:37:40.000
So, we do both of the versions, and we will end up getting t = 5.004 seconds and 9.423 seconds.
00:37:40.000 --> 00:37:44.600
In the last one, the problem with the falling rock, where he threw the rock down the cliff,
00:37:44.600 --> 00:37:48.300
there was only one answer that was true out of the two things that came out of it.
00:37:48.300 --> 00:37:51.300
But what about this one--is one of them wrong and the other one right?
00:37:51.300 --> 00:37:53.400
Is it only possible for one of them to happen first?
00:37:53.400 --> 00:37:58.500
Well, if that is the case--if we are only looking for what is the immediate time, the soonest time,
00:37:58.500 --> 00:38:04.000
when they are 90 meters away from each other, then it is 5.004 seconds.
00:38:04.000 --> 00:38:08.300
However, if we think about what is going on, let's try to visualize this.
00:38:08.300 --> 00:38:13.300
We have a car in the north and a car in the east.
00:38:13.300 --> 00:38:17.900
They start very far away from one another, but as they get closer and closer to one another,
00:38:17.900 --> 00:38:24.000
at some point, their distance...this one is going faster; this one is going slower; they are going to pass,
00:38:24.000 --> 00:38:31.000
so that their distance is close enough for it to be...at 5.04 seconds, they are now 90 meters away.
00:38:31.000 --> 00:38:33.800
Now, they pass, and they end up being very close briefly.
00:38:33.800 --> 00:38:40.100
But then, they keep going; and at some point on the reverse side, their distance begins to grow now, after they pass the intersection.
00:38:40.100 --> 00:38:41.700
So, they actually start to get farther away.
00:38:41.700 --> 00:38:50.400
After they pass the intersection, eventually it is going to be that they are now 90 meters away from each other, once again, at 9.423 seconds.
00:38:50.400 --> 00:38:54.600
And then, if they keep going on and on and on, they will never end up being 90 meters away from each other.
00:38:54.600 --> 00:39:00.400
But this is the first time they are 90 meters away, and then this is the second time.
00:39:00.400 --> 00:39:04.800
Now, it is quite likely that a question phrased in this way would only be asking for the first one.
00:39:04.800 --> 00:39:12.400
But we are actually able to find out both of the times that they are 90 meters away, assuming they maintain constant speeds and straight roads.
00:39:12.400 --> 00:39:15.700
That is pretty cool; all right, I hope you have a good sense of how to complete the square
00:39:15.700 --> 00:39:17.900
and how important and useful the quadratic formula can be.
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Make sure you memorize it; I know--I hate memorizing things, too; but it ends up coming up so often.
00:39:22.600 --> 00:39:28.500
You really have to have it memorized: [-b ± √(b² - 4ac)]/2a.
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Make sure you get that one burned in your memory, because it will show up a lot.
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And we will see you later; next time, we will talk more about the general nature of quadratics and parabolas
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and see how what we just did in this lesson will end up connecting to that one, as well.
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See you at Educator.com later--goodbye!