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Completing the Square and the Quadratic Formula
 In this lesson, we will be working just with quadratic polynomials: polynomials with degree 2. Thus, quadratics are of the form
where a, b, c are constant real numbers and a ≠ 0.ax^{2} + bx + c,  Whenever we take the square roots on both sides of an equation when doing algebra, we put a ± on one side. Example: x^{2} = 4 ⇒ x = ±2.
 To complete the square for any quadratic, we want to put it in the form
Once in this form, we can easily set it to 0 and solve.()^{2} − k.  We convert as follows:
Don't try to memorize the formula above, instead, watch the video and learn the general method behind it. While the formula will work, it's very difficult to remember. It's much easier to learn the stepbystep method to produce it.ax^{2} + bx + c = 0 ⇔ ⎛
⎝x+ b 2a⎞
⎠
= b^{2} − 4ac 4a^{2}
 From this conversion, we can create the quadratic formula: a formula that gives an easy way to solve for the roots of any quadratic polynomial.
x = −b ±
√b^{2}−4ac
2a
 To use the quadratic formula above, the polynomial must be set up in the format ax^{2} + bx + c = 0. The quadratic must be put into that format before you can use the formula.
 There are three possible numbers of roots for a quadratic to have: 2, 1, or 0. We determine this number from the discriminant (contained in the quadratic formula): b^{2}−4ac. This value tells us how many roots the polynomial has:
 b^{2} − 4ac > 0 ⇒ 2 roots;
 b^{2} − 4ac = 0 ⇒ 1 root;
 b^{2} − 4ac < 0 ⇒ 0 roots.
Completing the Square and the Quadratic Formula

 We could solve this problem by factoring like we learned in the last lesson, but it would be difficult to figure out the appropriate factors. We could also use the quadratic formula, but there's an easier way. Because the problem only has x^{2}'s and no x's, we can solve it simply by getting a number on one side of the equation and x^{2} on the other side. Then take the square root.
 Working through this, we find
2x^{2} + 16 = 50 ⇒ 2x^{2} = 34 ⇒ x^{2} = 17  Once we have it in this form, we can take the square root of both sides. ALWAYS REMEMBER, when you take the square root of both sides, you MUST put a ± sign (plusminus, saying that there are two versions at once: a positive version and a negative version).
x = ±
√17  Is it possible to simplify √{17} any more? No, because it has no factors other than itself, so we can't pull anything out of the square root.

 The goal of completing the square is to create something of the form (x + )^{2}. We do this by adding the appropriate number to each side so we factor it in such a way on the left side.
 Consider how (x + )^{2} expands. For any number in that blank (let's use r so we can explore it), we would have
Looking at this, we see to create something in the form (x + )^{2}, we need to take half of the number multiplying the x in our equation, then square it.(x+r)^{2} = x^{2} + 2r·x + r^{2}.  Simply put, to complete the square, halve the number on the x, square it, then add it to each side of the equation. In this case, we have −10x, so we will want to add
to each side of the equation.⎛
⎝−10 2⎞
⎠2
= (−5)^{2} = 25  Doing this, we have x^{2} − 10x +25 = 16, which we can factor:
(x−5)^{2} = 16  Now we can take the square root of both sides to obtain our answers. Don't forget to put a ± sign when you take the square root of both sides!
x−5 = ±4

 The goal of completing the square is to create something of the form (x + )^{2}. In this case, we're not working with an equation, so we can't just add things to both sides. That's alright, we'll see a way to deal with this later on.
 To complete the square, we first need to have there be no coefficient on the x^{2}, and we need to group the x with the x^{2}. We can do this by pulling out the coefficient in front of x^{2} from it and the x:
−4x^{2} − 24 x + 36 ⇒ −4(x^{2} + 6x) + 36  To complete the square of x^{2} + 6x, we need to add half of 6 squared: ([1/2] ·6)^{2} = 9. However, we have an issue: This is not an equation, so we can't just add a number to both sides. We can get around this though! Notice that nothing happens if we add 0, so we can add something that is equivalent to 0: 9 −9. As long as we can separate the 9 and the −9, we can use the 9 to complete the square.
 Add the 9−9 inside of the parentheses, next to the x^{2}+6x:
The only thing we want is x^{2}+6x+9, though. The −9 just gets in the way. Take it out of the parentheses by distributing the −4 on to it:−4(x^{2} + 6x+9−9) + 36. −4(x^{2} + 6x+9)+ 36 + 36.  At this point, we can now complete the square:
−4(x+3)^{2}+ 72  It can be easy to make a mistake, so it's a good idea to check your work. To be sure the answer is right, verify that
which turns out to be true.−4(x+3)^{2}+ 72 = −4x^{2} − 24x + 36,

 We need to see our equation in the same shape as ax^{2}+bx+c=0. It's already pretty close, but currently we're missing some stuff. We need to fill in this blank:
p^{2} + p − 20 = 0  Since there are no p's in the original equation, we have 0p. We now have p^{2} + 0p −20=0, which is closer to the format we want, but still not quite the same. To finish it up, we need to realize that p^{2} is the same thing as 1·p^{2} and that minus 20 can be written as plus −20.
 With those final things in mind, we have
and it is easy to identify a, b, and c. [Note: When you get used to using the quadratic formula, you don't have to take so many steps to figure out a, b, and c. Once you're comfortable, you can just glance at the equation and immediately figure them out.]1p^{2} + 0p + (−20) = 0,
 We're solving for all the x values that make the polynomial equation true. While we could do this with factoring, we're doing it with the quadratic formula. The quadratic formula says for ax^{2}+bx+c=0, we have
x = −b ±
√b^{2}−4ac2a.  For this problem, we have a=1, b=8, and c=−153. Plugging them in, we have
x = −8 ±
√8^{2}−4·1·(−153)2·1.  Simplifying, we have
x = −8 ±
√64+6122= −8 ±
√6762= −8 ±26 2= −4 ±13
 We're solving for all the t values that make the polynomial equation true. While we could do this with factoring, we're doing it with the quadratic formula. The quadratic formula says for equations of the form ax^{2}+bx+c=0, we have
x = −b ±
√b^{2}−4ac2a.  For this problem, we have a=−2, b=2, and c=24 and we're using t instead of x. Setting this up, we have
t = −2 ±
√2^{2}−4·(−2)·242·(−2).  Simplifying, we have
[Notice that ± and ± effectively mean the exact same thing here, because they are just saying that there are both plus and minus versions.]t = −2 ±
√4+192−4= −2 ±
√196−4= −2 ±14 −4= 1 2± 7 2
 We're solving for all the n values that make the polynomial equation true. While we could do this with factoring (although it would be very, very difficult), we're doing it with the quadratic formula. The quadratic formula says for equations of the form ax^{2}+bx+c=0, we have
x = −b ±
√b^{2}−4ac2a.  Before we can apply the quadratic formula, we first need to get it into that format: ax^{2}+bx+c=0. Right now, our equation looks totally different. To use the quadratic formula, we must expand, simplify, and get everything on one side so we have a 0 on the other side:
3n^{2}+21 − 5n = 3n − n^{2} + 44 ⇒ 4n^{2} − 8n − 23 = 0  We now have a=4, b=−8, and c=−23 and we're using n instead of x. Setting this up, we have
n = −(−8) ±
√(−8)^{2}−4·4·(−23)2·4.  Simplifying, we have
n = 8 ±
√64+3688= 8 ±
√4328= 8 ±12 √3 8= 1 ± 3√3 2

 The discriminant tells you how many roots/zeros a polynomial has. The quadratic formula is
but the discriminant is just the part underneath the square root: b^{2} − 4ac.x = −b ±
√b^{2}−4ac2a,  If the discriminant is greater than 0, there are 2 roots. If it equals 0, there is 1 root, and if it is less than 0, there are no roots at all.
 From the polynomial, we have a=4, b=−20, and c=25. Plugging them into the discriminant we have
Thus, the discriminant equals 0, so there is precisely 1 root.(−20)^{2} − 4 ·4 ·25 ⇒ 400 − 400 ⇒ 0  Because we're looking for how many ways the polynomial can equal 0, and we know there is only 1 root for the polynomial, then there is only one solution to the polynomial equation.

 The discriminant tells you how many roots/zeros a polynomial has. The quadratic formula is
but the discriminant is just the part underneath the square root: b^{2} − 4ac.x = −b ±
√b^{2}−4ac2a,  If the discriminant is greater than 0, there are 2 roots. If it equals 0, there is 1 root, and if it is less than 0, there are no roots at all.
 From the polynomial, we have a=2, b=7, and c=43. Plugging them into the discriminant we have
Thus, the discriminant equals −295, so there are no roots at all.7^{2} − 4 ·2 ·43 ⇒ 49 − 344 ⇒ −295

How long does it take for the ball to hit the ground?
 We must begin by understanding what the problem is asking. We have a function h(t) that tells us the height of a ball, and we've been asked to find out when the ball will hit the ground. To do this, we must realize that because h(t) is the height of the ball above the ground, when h(t) = 0, the ball will be touching the ground. This means we want to solve for what t will make h(t)=0.
 Plugging in h(t)=0 gives us an equation we can solve:
To solve this equation, we'll want to use the quadratic formula. The quadratic formula says for equations of the form ax^{2}+bx+c=0, we have0 = −4.9t^{2} + 7t + 11 x = −b ±
√b^{2}−4ac2a.  We have a=−4.9, b=7, and c=11. Setting this up, we have
t = −7 ±
√7^{2} − 4 ·(−4.9) ·112·(−4.9).  Simplifying, we get
t = −7 ±
√49 +215.6−9.8= −7 ±
√264.6−9.8  This means the quadratic formula gives us two possibilities for t:
t = −7 +
√264.6−9.8≈ −0.9456 and t = −7 +
√264.6−9.8≈ 2.3741  HOWEVER! It makes no sense for the ball to hit the ground at a negative time t. The function only applies after the ball is thrown, that is, in positive time t. Thus, the negative answer is extraneous, and we get rid of it, leaving us with only the positive answer.
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Completing the Square and the Quadratic Formula
Lecture Slides are screencaptured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
 Intro
 Introduction
 Square Roots and Equations
 Completing the Square: Motivation
 Polynomials that are Easy to Solve
 Making Complex Polynomials Easy to Solve
 Steps to Completing the Square
 Completing the Square: Method
 Solving Quadratics with Ease
 The Quadratic Formula
 Follow Format to Use Formula
 How Many Roots?
 The Discriminant
 Example 1: Complete the Square
 Example 2: Solve the Quadratic
 Example 3: Solve for Zeroes
 Example 4: Using the Quadratic Formula
 Intro 0:00
 Introduction 0:05
 Square Roots and Equations 0:51
 Taking the Square Root to Find the Value of x
 Getting the Positive and Negative Answers
 Completing the Square: Motivation 2:04
 Polynomials that are Easy to Solve
 Making Complex Polynomials Easy to Solve
 Steps to Completing the Square
 Completing the Square: Method 7:22
 Move C over
 Divide by A
 Find r
 Add to Both Sides to Complete the Square
 Solving Quadratics with Ease 9:56
 The Quadratic Formula 11:38
 Derivation
 Final Form
 Follow Format to Use Formula 13:38
 How Many Roots? 14:53
 The Discriminant 15:47
 What the Discriminant Tells Us: How Many Roots
 How the Discriminant Works
 Example 1: Complete the Square 18:24
 Example 2: Solve the Quadratic 22:00
 Example 3: Solve for Zeroes 25:28
 Example 4: Using the Quadratic Formula 30:52
Precalculus with Limits Online Course
Transcription: Completing the Square and the Quadratic Formula
Hiwelcome back to Educator.com.0000
Today, we are going to talk about completing the square and the quadratic formula.0002
In this lesson, we will be working just with quadratic polynomialsthat is, polynomials that have degree 2.0006
Quadratics are of the form ax^{2} + bx + c, degree 2; a, b, and c are constant real numbers, and a is not equal to 0.0011
Otherwise it wouldn't be a quadratic anymore, because we would have knocked out that x^{2}.0021
In the previous lesson, we talked all about finding the roots of polynomials.0025
Since quadratics appear often in nature, we have to find their roots a lot.0028
However, also from the previous lesson, we saw that finding roots and factors...it is not always an easy business to factor a polynomial.0031
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?0039
And it turns out that there is; this lesson is going to explore that method, which will make it easier for us.0045
First, let's remind ourselves of something we learned long ago in algebra: consider solving x^{2} = 1.0052
Now, our first automatic response would be to take the square root of both sides, and we might get x = 1.0057
But we have to remember that that is only have of the answer.0062
Hopefully we remember that, when we take the square root of both sides, we have to also introduce a plus and a negative version.0065
Remember, since (1)^{2} = 1, and 1^{2} = 1, we actually have two answers for this, x = 1 and x = 1.0071
When you square a negative, it loses its negativeness and becomes a positive number.0081
So, if we wanted, we could express this as x = ± 1, using this symbol right here, which we call the "plus/minus symbol."0085
The previous idea is given by this important rulewe have to always remember this any time we end up taking the square root.0094
Otherwise we will introduce mistakes: whenever we take the square roots on both sides of an equation,0100
when we are doing algebra, we put a plus/minus on one side.0105
So, for example, if we have x^{2} = 1, we take the square root of both sides; we get √(x^{2}) = ±√1,0109
at which point we get x = ± 1, which we can unfold into x = +1 and x = 1, our two answers.0116
These things might get us thinking, though; it is easy to solve equations that are in this form, x^{2} = k.0125
We get x = ±√k; so that is pretty easyif we could somehow get a quadratic to look like that, we would be doing pretty well.0132
So, what if we had a polynomial like x^{2}  16 = 0?0140
Well, then it is really easy: we just toss that 16 over; we get x^{2} = 16.0143
The square root of both sides: √(x^{2}) = ±√16; so we take the square root of both of those; x = ±4.0148
We managed to find the answerniceit worked really easily when we have this x^{2}  k = 0.0157
So, this method works great for x^{2}  k = 0, because we just move it over and get x^{2} = k,0166
at which point we take the square root and introduce that plus/minus.0174
But we couldn't find the roots of x^{2}  2x  3 with it, because we can't just move it over.0176
So, that is too bad...or could we?...maybe there is a way.0181
Let's say a little bird points out to us that x^{2}  2x  3 equals (x  1)^{2}  4.0186
Well, at this point, it is really easy to solve for the roots now.0195
We set x^{2}  2x  3 equal to 0, and then we use this piece of information right here.0197
We know that we can swap this and this; we have this right here, so we swap it out,0204
because we were told by that little bird (and we trust that little bird) that (x  1)^{2}  4 is the exact same thing.0210
So, at this point, we move our 4 over; we take the square root of both sides; √4 is 2, and we have ±,0216
because we have to introduce that plus/minus when we take the square root.0222
x  1 = ±2; we move the 1 over, and we get x = 1 ± 2, which is equal to 3 and 1.0225
We have gotten both of the answers for this quadratic.0233
Great; what if there was some way that we could do this for any quadratic?0236
If we could get this form of something squared minus k, it would be easy to find roots for any quadratic.0241
So, let's try to do this on 4x^{2} + 24x + 9 = 0.0248
We will see if we can find a method for...we will call it completing the square,0252
because we are going from this form where there is a bunch of stuff to this nice thing that is squared.0257
So, we will call it completing the square, because once we have a square, minus just a constant,0262
it is really, really easy to be able to solve for what the roots have to be.0266
So, let's move the 9 over first; we will get 4x^{2} + 24x = 9.0271
Now, how can we get 4x^{2} + 24x to become (_x + _)^{2}?0275
Well, there is that pesky 4, still in front of that x^{2}; so let's just start by getting rid of that 4.0281
We divide out the 4, and we have x^{2} + 6x = 9/4.0287
4x^{2}/4 becomes just x^{2}; 24x/4 becomes 6x (6 times 4 is 24); that equals 9/4; great.0292
From this format, we want to get some (x + _)^{2}, (x + r)^{2}.0302
Now, notice: (x + r)^{2} is equal to x^{2} + 2rx + r^{2}...0308
sorry, (x + r)^{2} becomes xr + rx; so we have two r's showing up on that x, 2r times x.0317
x^{2} + 2rx + r^{2}: we need to figure out what goes in these blanks.0326
We have a blank here; x^{2} + 6x + effectively a blank...to be able to complete and get this here.0330
So, (x + _)^{2}how can we do this? Well, we will use this information that we just had here.0337
We realize that x^{2} + 6x + 9 = (x + 3)^{2}.0343
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.0347
So, if 2r is 6, then 1r is 3; and we check this out: x^{2} + 6x + 9 = (x + 3)^{2}.0355
We check it: x times x is x^{2}; x times 3 is 3x, plus 3 times x is another 3x, so 6x; great; plus 3^{2}...3 times 3 is 9; great.0361
It checks out; so we have x^{2} + 6x = 9/4; that is what our equation was.0372
How do we get a 9 on the left? Simplewe just add a 9 to both sides.0377
So, add 9 to both sides, because we figured out that we want it to look like this.0381
Since we want it to look like this, we make it look like this through basic algebra manipulation.0386
Add 9 to both sides; we get x^{2} + 6x + 9 = 9/4 + 9.0390
The left side is now (x + 3)^{2}; we collapse it, and we have it equal to (9 + 36)/4,0394
since 9 is equal to 9 times 4, over 4, which equals 36/4.0402
It connects to that other 9/4 by getting a common denominator and then adding to it.0408
So, it is (9 + 36)/4; we simplify that, and we get (x + 3)^{2} = 27/4; great.0414
If we wanted to, we could easily solve this.0420
So, we take the square root of both sides: we get ±√27/√4 = x + 3easy.0422
We call this procedure, once again, completing the square, because we are going from a method0429
that doesn't really have this nice squared chunk to a thing that does have this nice squared chunk,0433
just minus some other factor or plus some other factor.0439
We can do this in general; we can do this to some general quadratic polynomial, ax^{2} + bx + c = 0.0443
We can do this in general, and basically follow the exact same method that we just did with numbers.0450
So, first we move the c over; just like we move the 9 over, we have c now.0455
Since we eventually want something of the form (x + _)^{2}, we don't want this pesky a getting in the way.0460
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 x^{2};0465
divide by a over here...c/a; great, so we get x^{2} + b/a(x) = c/a.0471
All right, next our goal is something of the form (x + _)^{2}.0478
Now, we notice, once again: (x + r)^{2}, whatever r is, equals x^{2} + 2rx + r^{2}.0484
Now, we want to match up to this format; we already have b/a(x), and we want 2r(x).0491
The x^{2} here matches with the x^{2} here; the 2rx here matches with the b/a(x) here;0501
and the r^{2} that we haven't introduced yet0508
is what the blank is that we don't know what we are going to put in yet.0510
So, if 2r is the same thing as b/a, if b/a = 2r, then that means b/2a = r.0514
So now, with that in mind, we know that what we want to introduce is r^{2}:0523
b/2a = r, so we want to add r^{2}, or (b/2a)^{2}, to both sides.0527
So, we add r^{2} = b^{2}/4a^{2}, and we complete the square.0535
x^{2} + b/a(x) + b^{2}/4a^{2}...that collapses into (x + b/2a)^{2}.0539
Check that out really quickly: x times x becomes x^{2}; great; x times b/2a becomes b/2a(x),0547
plus it will happen a second time; so b/2a + b/2a becomes 2b/2a, so just b/a, still times x;0556
b/2a times b/2a becomes b^{2}, over...2 times 2 is 4...a times a is squared; so it is b^{2}/4a^{2}.0564
Great; that checks out; and we added this b^{2}/4a^{2} to both sides; we can't just add it to one side.0571
And so, that will collapse into (b^{2}  4ac)/4a^{2}, because we have c/a; so that becomes 4ac/4a^{2}.0577
So, we can get them on a common denominator; so we have completed the square.0589
Great; that is a general form.0592
At this point, we have shown that any equation that starts as ax^{2} + bx + c = 0 is equivalent,0597
through completing the square, to this equation right here.0603
It is a little bit complex, but we just proved that we can just do that through basic algebraic manipulation.0606
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.0612
For example, if we have 4x^{2} + 8x + 2, then our a is equal to 4; our b equals 8; and our c equals 2.0618
Oh, let's colorcode that; so a at 4 is red; b at 8 is blue; and green is c = 2; lovely.0627
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.0637
So, we follow this format; and we have blue 8^{2} here; our a's...4 here and 4 here; 4 here;0650
and then finally, our green is here, and this coefficient here just stays here; this coefficient here just stays here;0666
this coefficient here just stays here; so if we wanted to, at this point, we could just do some arithmetic;0675
and we would be able to simplify that, and then we would be able to take the square root,0680
and basically just be able to solve for it, and we would be able to get the answer.0683
But we can go one step farther, and we can just set up a general formula to solve any quadratic, ax^{2} + bx + c = 0.0686
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.0693
So, at this point, we have shown that ax^{2} + bx + c = 0 is the same thing;0700
it is equivalent to the completedsquare version of (x + b/2a)^{2} = (b^{2}  4ac)/4a^{2}.0705
So, we just take the square root of both sides to get to the x: x + b/2a = ±√(b^{2}  4ac...0713
what is the square root of 4a^{2}? 4 comes out as 2; a^{2} comes out as a.0719
So, we get the 2a on the bottom; so (x + b/2a) = ±√(b^{2}  4ac), all over 2a.0724
Next, we isolate for x; we subtract the b/2a, plus or minus √(b^{2}  4ac), all over 2a.0733
Look, they are already in common denominators; so we get x = [b ± √(b^{2}  4ac)]/2a.0740
We have the quadratic formula, an easy way to solve for the roots of any quadratic polynomial.0750
So, as long as we have some quadratic polynomial like this, we just plug into this thing and do some arithmetic.0756
It might get ugly; it might require a calculator; it might not be really easy arithmetic.0761
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.0766
We just plug in and go, and an answer will pop out.0772
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.0775
But the quadratic formula is going to come up so often that you are going to end up needing to see0780
this [b ± √(b^{2}  4ac)]/2a...I am going to have to recommend that you probably want to memorize this thing.0785
Memorize the quadratic formula, because it will show up a lot.0793
And even if your teacher doesn't absolutely require you to have it memorized in another class,0797
you are going to end up seeing this so often, and you are going to have to solve for so many quadratics,0802
that you want to just have this ready, so that you can pull it out any time.0807
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."0810
It comes up a lot, so it is good to just have it memorized.0816
All right, follow the format if you are going to use the formula.0819
It is really important to note that, if you want to use the quadratic formula,0822
the polynomial must be set up in this format, ax^{2} + bx + c = 0.0825
It absolutely has to be set up in this format.0830
For example, if we have 2x^{2}  47x + 23, then our a equals 2; our b equals 470831
(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 23.0840
So, a = 2; b = 47; c = 23; great.0849
But it would be totally wrong, absolutely wrong, to say x^{2} + 3x  4 = 2x + 8 gives us a = 1, b = 3, c = 4.0857
We have this equals...stuff over here; it has to equal 0; otherwise it doesn't work.0867
It has to be in this format of ax^{2} + bx + c = 0.0873
That is how we derive completing the square; that is how we derive the quadratic formula.0877
If it is not in this format, it breaks down entirely; we can't use the quadratic formula.0882
So, we have to put it into this format before we can use the quadratic formula.0886
It absolutely has to be in the form; otherwise, it just doesn't work.0890
How many roots does a quadratic have?0894
Now, of course, not all quadratics have the same number of roots.0896
The graphs below show the three possibilities: we have one where it intersects it twice (one here, one heretwo roots);0899
we have it where it intersects it just once (it barely grazes and touches, barely just hitting it once);0908
or we have absolutely none, where it never manages to cross the xaxis.0914
And of course, these could all flip the other way; we could have it going down this way.0919
We could have it barely touching on this side; and we could also have it crossing over like this.0924
There is no guarantee that it has to be pointed or look like these.0931
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,0933
or cut not at all, because it never manages to cross it.0940
So, the quadratic formula actually manages to show us which one of these situations we are in.0943
The way we do this is through the discriminant.0948
Remember the formula: b ± √(b^{2}  4ac), all over 2a.0950
You are going to hear that a lot; it is good to memorize.0955
The expression b^{2}  4ac is the discriminant; it tells us how many roots there are.0957
b^{2}  4ac being greater than 0 means that there are two roots.0963
If it is equal to 0, then we have one root; and if it is less than 0, we have 0 roots,0971
each corresponding to those colors on the last picture, as well.0977
So, the discriminant tells us how many rootswhat kind of situation we are in0981
for how often our parabola is going to actually manage to cross over that xintercept and touch that xaxis.0984
Why or how does it workwhat is going on here?0990
Let's look at the quadratic formula once again.0992
Remember: our discriminant is the b^{2}  4ac part, the part underneath the square root.0995
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.1000
We have this plus or minus here; so if b^{2}  4ac is positive, two are going to come out, because of the plus or minus.1007
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).1014
So, we have a plus version and a minus version; that is two different worlds, so we get two different answers.1021
But if we have b^{2}  4ac equals 0, then ± 0 is just 0.1028
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.1033
So, we just have one rootonly one possibility.1042
Finally, if b^{2}  4ac is negative (that is, less than 0), the square root fails entirely, so there are no answers.1045
It is impossible to take the square root of a negative number, remember, because any positive squared becomes positive;1052
any negative squared becomes negative; so there is no number out there that, when you square it, will become a negative.1058
So, there are no answers if b^{2}  4ac is negativeat least, there are no real answers.1064
We will talk about how things get a little shady once we get into complex numbers.1068
But for right now, the discriminant tells us that there are no answers if b^{2}  4ac is negative.1071
Really, we can just look at how this interacts with square roothow many things can come out of ± square root.1075
If two things can come out, because it is a positive number under the square root, then we have two possibilities, two answers.1083
If 0 is under the square root, then there is one possibility, because it is just one possibility underneath √0.1089
So, ±0 is just one thing; if it is ± the square root of a negative number,1097
then that is impossible to do in the first place, so we have no possibilities under it; we have no answers.1100
All right, we are ready for some examples.1105
Complete the square on the polynomial 3x^{2}  30x + 87 to give an equivalent expression.1107
Now, we didn't formally talk about completing the square when it wasn't equal to 0.1111
But we can follow the exact same method: 3x^{2}  30x + 87the first thing we do is...we want to look at that 87 as being off on the side.1115
It is still part of the expression, but it is not what we really want to work with, fundamentally.1125
Now, we have a 3 in here, so we want to pull this 3; we will pull it out; we get 3(x^{2} ...30...pulling out a 3 gets 10x) + 87 (is still over there).1128
3(x^{2}  10x); we checkyes, it checks out; we still have the same thing there.1140
The next step: we want to figure out + _...what blanks can go in there?1145
Well, remember: if it was (x + r)^{2}, then that would become x^{2} + 2rx + r^{2}.1150
So, what makes up our 2r right now? Our 2r here is our 10.1158
So, if 2r = 10, then our r by itself would be equal to 5; so we want to get a 25 inside.1162
So, we have 3(x^{2}  10x +...[we want r^{2} equal to]25); 5 times 5 is 25, so it is +25 on the inside.1171
So, we want that to show up; of course, if we just change our expression aroundwe just put a number in there1184
it is not the same expression; we just broke our mathematicsit doesn't work like that.1189
So, we have to make sure that however much we put in on one place, we take out of somewhere else.1192
I can have any number...5...and I could add 3 and subtract 3, and it would have no effect.1197
I would still be left with 5, because the same thing is adding 0.1202
So, if we add 25 into the inside of our quantity, how much do we need to take away on the outside?1205
Well, putting in 25 on the inside...remember, it is 3 times (....+ 25).1211
Well, that is going to be ..... + 75; 3 times 25 goes to connect like that; 2 times 25 means we need to take away 751218
on the outside to keep our scale balanced; we put 25 into the inside of the parentheses;1233
3 times 25 is 75; so we need to take 75 out.1240
We put a total of 75 in the expression; so if we take a total of 75 out, our scale remains balanced.1243
 75...and then we still have to bring on what was in the expression before, + 87.1249
So, 3(x^{2}  10x + 25)...(x + r)^{2}...our r equals 5, so we have (x  5)^{2}.1255
Let's check and make sure that is still correct: x^{2}, x times x, x^{2}.1264
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.1268
Minus 75 + 87; that becomes + 12; and now we have something that is equivalent.1277
3(x  5)^{2} + 12...let's check to make sure that is correct.1287
3(x  5)^{2} would become x^{2}  10x + 25, plus 12; 3x^{2}  30x + 75 + 12; 3x^{2}  30x + 87.1293
Greatit checks out; that is the same thing as what we started with.1311
All right, the next onethe second example: Solve x^{2} + 10x  20 = 4x  16.1314
So, we have that nice, fancy quadratic formula; let's try it out.1320
The first thing, though: it is not currently equal to 0it is not set to 0,1324
so we need to get the whole thing so it looks like that ax^{2} + bx + c = 0.1329
So, let's move things around: we subtract 4x; we add 16; we get x^{2} + 6x  4 = 0.1333
Great; so we are now set upwe have a = 1, b = 6, c = 4; we are in that format.1348
Our normal format is ax^{2} + bx + c = 0; so we have that parallel.1361
What is our formula? The roots are going to be x = [b ± √(b^{2}  4ac)]/2a.1367
All right, so we start plugging into that: we have x =...plug in our blue 6, our b; plus or minus the square root...1382
b^{2} is 6^{2}; minus 4 times a (1) times c (4)...keep that going;1397
the colors are getting a little bit crazy here, but the basic idea going on is still the same;1410
6 ± √...2 times...and the red one here, a; and notice these coefficientsthey just stick around the whole time; 4, 4.1417
They just stick around no matter what; the ± moves down; the negative moves down; so those things are always there.1429
At this point, all we have to do is solve it out; so x = [6 ± √(36...(4)(1)(4)...1436
two of those cancel out, but we are still left with a negative, so... 16, over 2 times...1447
we are going to replace that with what it should become, 2 times 1; I got confused by all the colors.1455
2 times 1 becomes 2; so we have x = 6/2 ± √(20)/2.1461
x =...this becomes positive 3, plus or minus the square root of 20 (is equal to √(4)(5)); we get 2√5.1475
So, we replace that down here; so we have 3 ± 2√5, all over 2.1491
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,1499
and the negative to become a plus, so it didn't really do anything: plus/minus is the same thing as minus/plus.1505
We are basically where we were before: √5.1510
So, our answers are going to be x = 3  √5 and 3 + √5; those are the solutions to that,1513
because we found the roots to when we turn it into the format that we could use it on; great.1522
A man standing on the top of a 127metertall cliff throws a ball directly down at 10 meters per second.1530
The height of the ball above the ground is given by: height at time t equals 4.9t^{2}  10t + 127, where t is in seconds.1536
How long does it take for the ball to hit the ground?1546
We have this man; he is standing on top of a cliff; and for some reason, he throws a ball down.1548
All right, the ball is going down; it is moving down towards the ground down here.1555
Let's see what this means: what is ground?1561
Ground is h =...what does that mean?1567
Well, we notice that if we plug in 0, then that is going to be just as he threw it, which would be...1570
the 0's would cancel out; the t would cancel with the t here;1574
and we would be left with just 127, which is where he starts at, 127 meters high.1576
So, that makes sensethat the ground is going to be 0 meters high; it makes intuitive sense.1583
So, we know that what we are looking for is when the ball hits the ground.1587
When does the ball hit the ground? The ball hits the ground at h = 0.1591
That is what the height we are looking for is: ground is normally put at h = 0.1599
You can sometimes move it around; but for the most part, you will end up seeing,1603
in any word problem where they are talking about the height, that the ground level1606
and normally, whatever our base level is consideredis a height of 0.1608
How long does it take for the ball to hit the ground?1613
Well, that means, if we are looking for when h equals 0, that we are going to use that in here in our functions.1615
So, 0 = 4.9t^{2}  10t + 127; we plug this into our quadratic formula.1620
We are not looking at x anymore; we are looking at when t is going to give our roots.1631
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 4.9.1634
t = +10 ± √(100 ...that will end up...we work that out with a calculator...2489.2), over 9.8.1659
So, at this point, that is not super easy to work out; so we are going to start plugging into a calculator.1674
We won't actually do that here; but we are going to basically plug two different expressions into our calculator.1679
The plus version is...one mistake there...that was 4 times 4.9, so it becomes +.1683
Otherwise it would be impossible, because we would have a negative underneath our square root.1697
That is what made me catch that.1700
2589.2 divided by 9.8...and 10 plus 10 minus √2589.2/9.8.1702
We plug that into our calculator, and we get two different answers: t = 6.213 and 4.172.1717
Now, at first, that should set off some alarm bells in our head.1728
We throw a ball down a cliff...we imagine this in our head; that is the very first stepwe imagine it in our head.1731
You throw a ball down some tall height, and eventually it hits the ground.1736
All right, that is the end; it doesn't hit the ground at two different times.1739
And so, what does this mean?we have two different answers here, so which one is the correct answer, 6.213 or 4.172?1743
Which one is right? We think; we know that this is t given in seconds, so we know what happens.1751
He throws the ball down, and so forward in time, it is falling; we are going by this equation.1758
But what about a negative timedid he throw the ball before 0 seconds?1764
No, he throws the ball at 0 seconds; it starts at his height at 0 seconds.1768
So, it must be that h(t) is only trueits domain is only 0 to positive infinity.1773
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.1781
So, h(t) is only true from t = 0 until the ball hits the grounduntil whatever t the ball hits the ground at.1788
So, that is sort of an implicit thing that we hadn't explicitly stated; but we had to understand what is going on,1799
because otherwise we will get two answers, and one of them is going to be wrong.1804
We have to realize what is going on; we don't want to just blindly do what the formula told us.1807
We want to think about what this represents; word problems require thinking.1812
So, 6.213 and 4.172...we realize it is only true from t = 0 to higher numbers, until the ball hits the ground.1816
The ball hits the ground at 4.172; and 6.213 is an extraneous solutionit is impossible to look at those times,1823
because that is back before this function ever ended up even being used.1831
The function comes into existence only once the ball is thrown at time t = 0.1837
So, negative times are completely extraneouswe can't use those answers.1842
And we get 4.172 seconds as the correct answer; great.1846
The final example: Two cars are approaching a rightangle intersection on straight roads.1853
The first one is coming from the north at a constant speed of 30 meters per second,1857
while the second one is from the east at a constant speed of 25 meters per second.1861
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?1865
This is the classic nightmare word problem with so many things herewhat are we going to do?1871
Well, we just start figuring out what it is telling us; then we will work on the math.1876
So, the first thing we do is try to make a picture of a rightangle intersection.1880
We know what an intersection looks like on the street.1884
Two streets intersect one another; we know that they came together at a right angle, because it says "rightangle intersection."1888
Great; they are on straight roads, so we are guaranteed the fact that straight lines come out of it; so this makes sense.1894
The first one is coming from the north; well, let's put north as being this way.1899
This car is up here; it is coming from the north at a constant speed of 30 meters per second.1904
It is going down 30 meters per second; where is it right now?1909
We know it is 200 meters away at the start; what about the other one?1913
The other one is coming from the east (east will be over here), and it is going at 25 meters per second.1919
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.1928
How much time is there until they have a distance of 90 meters between them?1937
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."1940
If we have a point here and a point here, then the distance is just the distance between those two points.1947
But we don't have any information about the distance between them; we are not told how far away they are.1955
But we are told how far they are from this intersection in the middle.1959
If we have...here is x and here is y...look, we can use the Pythagorean theorem: x^{2} + y^{2} = d^{2}.1964
Great; so we have a way of relating these two things.1973
But if we are just frozen at 200, 200^{2} + 200^{2} = distance squared,1976
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.1980
We also have to have a way of their movement, their motion, affecting this.1985
So, they start at 200 away; but then they start to get closer and closer to that intersection.1988
It is going to be 200  30t, because it is going to be that, as they get closer to the intersection,1993
as more time goes on, more of their distance from the intersection will disappear.2000
So, the 200  30t...and then, the other one is 200  25t, because the red one, the north car,2004
is coming at 30 meters per second; so for every second that goes by, it will have moved 30 meters2014
towards the intersection, so it will be 200  30t; and the blue car, the east car, will be 200  25t,2019
because for every second, it moves 25 meters towards the intersection: 200 less 25 meters.2026
OK, so at this point, we have a real understanding of what is going on.2034
We can now put this d in here; and we can connect all of these ideas.2036
So, we have 200  30t is what describes the red car, our north car.2043
And 200  25t is what describes the east car, our blue car.2049
And we are looking for when the distance is equal to 90.2055
So remember: we had x^{2} + y^{2} = d^{2} from the fact that this is just a nice, normal Pythagorean triangle.2060
We can use the Pythagorean theorem here: the square of the two sides is equal to the square of the hypotenuse.2068
So, we have 90^{2} = (200  30t)^{2} (our red carour north car) + (200  25t)^{2} (our east carour blue car); great.2075
And now we are trying to solve this and figure out when t is going to make this true.2086
At what t will that equation there be true?2089
So, we just start working it out; now, this is going to get pretty big pretty fast, because we have big numbers.2092
But luckily, we have access to calculators in this world.2097
So, 90^{2}...plug that in; that comes out as 8100; 200 times 200 becomes 40000; minus 30t...2099
200 times 30t, plus 30t times 200; 200 times 30t is 6000, but we have to double that,2110
so it is 12000t; minus 30t times 30t...+ 900 t^{2}.2116
The next one is + 40000 again for the other portion; 200 times 25 becomes 5000.2126
But then, we also have to have 25 times 200 again the other way, so it is 500 doubled, so 10000t.2135
Plus 25 times 25, so + 625t^{2}...2144
All right, we simplify this; this looks like something that could eventually turn into a quadratic.2152
So, we say, "Oh, it is time to use the quadratic formula!"2158
So, we need to get into that form: we will subtract 8100 over, so 8100, 8100 over here...2160
We look at 625t^{2} + 900t^{2}; we get 1525t^{2}.2170
Next, 10000t; 12000t; 22000t; 40000 and 40000...minus 81000, plus 71900; wow.2179
But we are in a position where we can now use the quadratic formula.2193
Once again, it is a good thing that we have calculators; otherwise this would be really difficult.2195
But we can use the quadratic formula now.2199
So, t = b, (22000), so 22000, plus or minus the square root of b^{2}, 22000^{2};2201
we will drop the negative sign, just because it is going to get squared anyway; minus 4, times 1525, times 71900.2213
At this step, we might toss those parts in, just the part underneath the square root, use the discriminant,2225
and make sure that there is an answer; it will turn out that there is an answer, so we will just keep going.2230
And that is going to be divided by 2 times 1525.2234
I won't work this all out here; I am going to trust the fact that you can do the two different versions.2239
Remember: there is a plus version, and then there is a minus version.2243
So, we do both of the versions, and we will end up getting t = 5.004 seconds and 9.423 seconds.2247
In the last one, the problem with the falling rock, where he threw the rock down the cliff,2260
there was only one answer that was true out of the two things that came out of it.2264
But what about this oneis one of them wrong and the other one right?2268
Is it only possible for one of them to happen first?2271
Well, if that is the caseif we are only looking for what is the immediate time, the soonest time,2273
when they are 90 meters away from each other, then it is 5.004 seconds.2278
However, if we think about what is going on, let's try to visualize this.2284
We have a car in the north and a car in the east.2288
They start very far away from one another, but as they get closer and closer to one another,2293
at some point, their distance...this one is going faster; this one is going slower; they are going to pass,2298
so that their distance is close enough for it to be...at 5.04 seconds, they are now 90 meters away.2304
Now, they pass, and they end up being very close briefly.2311
But then, they keep going; and at some point on the reverse side, their distance begins to grow now, after they pass the intersection.2314
So, they actually start to get farther away.2320
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.2322
And then, if they keep going on and on and on, they will never end up being 90 meters away from each other.2330
But this is the first time they are 90 meters away, and then this is the second time.2334
Now, it is quite likely that a question phrased in this way would only be asking for the first one.2340
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.2345
That is pretty cool; all right, I hope you have a good sense of how to complete the square2352
and how important and useful the quadratic formula can be.2356
Make sure you memorize it; I knowI hate memorizing things, too; but it ends up coming up so often.2358
You really have to have it memorized: [b ± √(b^{2}  4ac)]/2a.2362
Make sure you get that one burned in your memory, because it will show up a lot.2368
And we will see you later; next time, we will talk more about the general nature of quadratics and parabolas2373
and see how what we just did in this lesson will end up connecting to that one, as well.2378
See you at Educator.com latergoodbye!2382
2 answers
Last reply by: Professor SelhorstJones
Sun Nov 8, 2015 4:27 PM
Post by Peter Ke on November 7, 2015
Hello, I am just curious cause I think I forgot. When square rooting why do you need both the positive and the negative?
For example, x^2 = 1 which is:
x = 1
x = 1
and not just x = 1 only.