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Lecture Comments (3)

2 answers

Last reply by: Professor Selhorst-Jones
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.

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
    ax2 + bx + c,
    where a, b, c are constant real numbers and a ≠ 0.
  • Whenever we take the square roots on both sides of an equation when doing algebra, we put a ± on one side. Example:  x2 = 4   ⇒   x = ±2.
  • To complete the square for any quadratic, we want to put it in the form
    ()2 − k.
    Once in this form, we can easily set it to 0 and solve.
  • We convert as follows:
    ax2 + bx + c = 0     ⇔    
    x+ b


    = b2 − 4ac


    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 step-by-step method to produce it.
  • 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 ±





  • To use the quadratic formula above, the polynomial must be set up in the format ax2 + 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): b2−4ac. This value tells us how many roots the polynomial has:
    • b2 − 4ac > 0  ⇒   2 roots;
    • b2 − 4ac = 0  ⇒   1 root;
    • b2 − 4ac < 0  ⇒   0 roots.

Completing the Square and the Quadratic Formula

Solve the below equation for all possible values of x:
2x2 + 16 = 50
  • 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 x2's and no x's, we can solve it simply by getting a number on one side of the equation and x2 on the other side. Then take the square root.
  • Working through this, we find
    2x2 + 16 = 50     ⇒     2x2 = 34     ⇒     x2 = 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 (plus-minus, saying that there are two versions at once: a positive version and a negative version).
    x = ±


  • 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.
x = ±√{17}, or, equivalently, x = √{17} and x = − √{17}
Complete the square to find the solutions to
x2 − 10x = −9.
  • 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
    (x+r)2 = x2 + 2r·x + r2.
    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.
  • 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




    = (−5)2 = 25
    to each side of the equation.
  • Doing this, we have x2 − 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
x = 5 ±4 or, equivalently, x=9 and x=1
Complete the square on the below polynomial to obtain an equivalent expression:

−4x2 − 24x + 36
  • 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 x2, and we need to group the x with the x2. We can do this by pulling out the coefficient in front of x2 from it and the x:
    −4x2 − 24 x + 36     ⇒     −4(x2 + 6x) + 36
  • To complete the square of x2 + 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 x2+6x:
    −4(x2 + 6x+9−9) + 36.
    The only thing we want is x2+6x+9, though. The −9 just gets in the way. Take it out of the parentheses by distributing the −4 on to it:
    −4(x2 + 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
    −4(x+3)2+ 72 = −4x2 − 24x + 36,
    which turns out to be true.
−4(x+3)2+ 72
To use the quadratic formula, a polynomial equation must be set up in the format ax2 + bx+c = 0. Identify a, b, and c for the polynomial equation

p2−20 = 0.
  • We need to see our equation in the same shape as ax2+bx+c=0. It's already pretty close, but currently we're missing some stuff. We need to fill in this blank:
    p2 +  p − 20 = 0
  • Since there are no p's in the original equation, we have 0p. We now have p2 + 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 p2 is the same thing as 1·p2 and that minus 20 can be written as plus −20.
  • With those final things in mind, we have
    1p2 + 0p + (−20) = 0,
    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.]
a=1, b=0, c=−20
Using the quadratic formula, solve x2+8x−153=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 ax2+bx+c=0, we have
    x =
    −b ±


  • For this problem, we have a=1, b=8, and c=−153. Plugging them in, we have
    x =
    −8 ±


  • Simplifying, we have
    x =
    −8 ±


    −8 ±


        =     −8 ±26

        =     −4 ±13
x=−4 ±13 or, equivalently, x=−17 and x=9
Using the quadratic formula, solve −2t2 + 2t + 24=0.
  • 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 ax2+bx+c=0, we have
    x =
    −b ±


  • 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 ±


  • Simplifying, we have
    t =
    −2 ±


    −2 ±


        =     −2 ±14

        =     1

    ± 7

    [Notice that ± and ± effectively mean the exact same thing here, because they are just saying that there are both plus and minus versions.]
t=[1/2] ±[7/2] or, equivalently, t=−3 and t=4
Using the quadratic formula, solve 3(n2+7) − 5n = 3n − n2 + 44.
  • 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 ax2+bx+c=0, we have
    x =
    −b ±


  • Before we can apply the quadratic formula, we first need to get it into that format: ax2+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:
    3n2+21 − 5n = 3n − n2 + 44     ⇒     4n2 − 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) ±


  • Simplifying, we have
    n =
    8 ±


    8 ±


        =     8 ±12 √3

        =     1 ± 3√3

n=1 ±[(3√3)/2] or, equivalently, n=1 + [(3√3)/2] ≈ 3.598 and n=1 − [(3√3)/2] ≈ −1.598
Using the discriminant, determine how many solutions exist to the polynomial equation
4x2 − 20x +25=0.
  • The discriminant tells you how many roots/zeros a polynomial has. The quadratic formula is
    x =
    −b ±


    but the discriminant is just the part underneath the square root: b2 − 4ac.
  • 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
    (−20)2 − 4 ·4 ·25     ⇒     400 − 400     ⇒     0
    Thus, the discriminant equals 0, so there is precisely 1 root.
  • 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.
One solution
Using the discriminant, determine how many roots exist for the function
f(w) = 2w2 +7w +43.
  • The discriminant tells you how many roots/zeros a polynomial has. The quadratic formula is
    x =
    −b ±


    but the discriminant is just the part underneath the square root: b2 − 4ac.
  • 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
    72 − 4 ·2 ·43     ⇒     49 − 344     ⇒     −295
    Thus, the discriminant equals −295, so there are no roots at all.
The function has no roots
A ball is thrown upward out of window at a speed of 7 m/s. The window is at a height of 11 m. The height of the ball above the ground can be given by

h(t) = −4.9t2 + 7t + 11,
where t is the number of seconds after the ball has been thrown.
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:
    0 = −4.9t2 + 7t + 11
    To solve this equation, we'll want to use the quadratic formula. The quadratic formula says for equations of the form ax2+bx+c=0, we have
    x =
    −b ±


  • We have a=−4.9, b=7, and c=11. Setting this up, we have
    t =
    −7 ±

    72 − 4 ·(−4.9) ·11

  • Simplifying, we get
    t =
    −7 ±

    49 +215.6

    −7 ±


  • This means the quadratic formula gives us two possibilities for t:
    t =
    −7 +


    ≈ −0.9456        and        t =
    −7 +


    ≈ 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.
The ball hits the ground after approximately 2.3741 s

*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.


Completing the Square and the Quadratic Formula

Lecture Slides are screen-captured 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 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

Transcription: Completing the Square and the Quadratic Formula

Hi--welcome back to

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 polynomials--that is, polynomials that have degree 2.0006

Quadratics are of the form ax2 + bx + c, degree 2; a, b, and c are constant real numbers, and a is not equal to 0.0011

Otherwise it wouldn't be a quadratic anymore, because we would have knocked out that x2.0021

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 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 x2 = 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 12 = 1, we actually have two answers for this, x = 1 and x = -1.0071

When you square a negative, it loses its negative-ness 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 rule--we 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 x2 = 1, we take the square root of both sides; we get √(x2) = ±√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, x2 = k.0125

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.0132

So, what if we had a polynomial like x2 - 16 = 0?0140

Well, then it is really easy: we just toss that 16 over; we get x2 = 16.0143

The square root of both sides: √(x2) = ±√16; so we take the square root of both of those; x = ±4.0148

We managed to find the answer--nice--it worked really easily when we have this x2 - k = 0.0157

So, this method works great for x2 - k = 0, because we just move it over and get x2 = k,0166

at which point we take the square root and introduce that plus/minus.0174

But we couldn't find the roots of x2 - 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 x2 - 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 x2 - 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 4x2 + 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 4x2 + 24x = -9.0271

Now, how can we get 4x2 + 24x to become (_x + _)2?0275

Well, there is that pesky 4, still in front of that x2; so let's just start by getting rid of that 4.0281

We divide out the 4, and we have x2 + 6x = -9/4.0287

4x2/4 becomes just x2; 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 x2 + 2rx + r2...0308

sorry, (x + r)2 becomes xr + rx; so we have two r's showing up on that x, 2r times x.0317

x2 + 2rx + r2: we need to figure out what goes in these blanks.0326

We have a blank here; x2 + 6x + effectively a 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 x2 + 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: x2 + 6x + 9 = (x + 3)2.0355

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.0361

It checks out; so we have x2 + 6x = -9/4; that is what our equation was.0372

How do we get a 9 on the left? Simple--we 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 x2 + 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 + 3--easy.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, ax2 + 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 x2;0465

divide by a over here...-c/a; great, so we get x2 + 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 x2 + 2rx + r2.0484

Now, we want to match up to this format; we already have b/a(x), and we want 2r(x).0491

The x2 here matches with the x2 here; the 2rx here matches with the b/a(x) here;0501

and the r2 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 r2:0523

b/2a = r, so we want to add r2, or (b/2a)2, to both sides.0527

So, we add r2 = b2/4a2, and we complete the square.0535

x2 + b/a(x) + b2/4a2...that collapses into (x + b/2a)2.0539

Check that out really quickly: x times x becomes x2; 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 b2, over...2 times 2 is 4...a times a is squared; so it is b2/4a2.0564

Great; that checks out; and we added this b2/4a2 to both sides; we can't just add it to one side.0571

And so, that will collapse into (b2 - 4ac)/4a2, because we have -c/a; so that becomes -4ac/4a2.0577

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 ax2 + 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 4x2 + 8x + 2, then our a is equal to 4; our b equals 8; and our c equals 2.0618

Oh, let's color-code 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 82 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, ax2 + 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 ax2 + bx + c = 0 is the same thing;0700

it is equivalent to the completed-square version of (x + b/2a)2 = (b2 - 4ac)/4a2.0705

So, we just take the square root of both sides to get to the x: x + b/2a = ±√(b2 - 4ac...0713

what is the square root of 4a2? 4 comes out as 2; a2 comes out as a.0719

So, we get the 2a on the bottom; so (x + b/2a) = ±√(b2 - 4ac), all over 2a.0724

Next, we isolate for x; we subtract the b/2a, plus or minus √(b2 - 4ac), all over 2a.0733

Look, they are already in common denominators; so we get x = [-b ± √(b2 - 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 ± √(b2 - 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, ax2 + bx + c = 0.0825

It absolutely has to be set up in this format.0830

For example, if we have 2x2 - 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 x2 + 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 ax2 + 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 here--two 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 x-axis.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 ± √(b2 - 4ac), all over 2a.0950

You are going to hear that a lot; it is good to memorize.0955

The expression b2 - 4ac is the discriminant; it tells us how many roots there are.0957

b2 - 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 roots--what kind of situation we are in0981

for how often our parabola is going to actually manage to cross over that x-intercept and touch that x-axis.0984

Why or how does it work--what is going on here?0990

Let's look at the quadratic formula once again.0992

Remember: our discriminant is the b2 - 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 b2 - 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 b2 - 4ac equals 0, then ± 0 is just 0.1028

So either way, it is just the same world; √ or minus doesn't really matter if we go with the plus or the minus; we get the same thing.1033

So, we just have one root--only one possibility.1042

Finally, if b2 - 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 b2 - 4ac is negative--at 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 b2 - 4ac is negative.1071

Really, we can just look at how this interacts with square root--how 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 3x2 - 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: 3x2 - 30x + 87--the 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(x2 -...30...pulling out a 3 gets 10x) + 87 (is still over there).1128

3(x2 - 10x); we check--yes, 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 x2 + 2rx + r2.1150

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(x2 - 10x +...[we want r2 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 around--we just put a number in there--1184

it is not the same expression; we just broke our mathematics--it 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(x2 - 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: x2, x times x, x2.1264

x times -5 is -5x; -5 times x is -5x again, so double that: 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 x2 - 10x + 25, plus 12; 3x2 - 30x + 75 + 12; 3x2 - 30x + 87.1293

Great--it checks out; that is the same thing as what we started with.1311

All right, the next one--the second example: Solve -x2 + 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 0--it is not set to 0,1324

so we need to get the whole thing so it looks like that ax2 + bx + c = 0.1329

So, let's move things around: we subtract 4x; we add 16; we get -x2 + 6x - 4 = 0.1333

Great; so we are now set up--we have a = -1, b = 6, c = -4; we are in that format.1348

Our normal format is ax2 + bx + c = 0; so we have that parallel.1361

What is our formula? The roots are going to be x = [-b ± √(b2 - 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

b2 is 62; 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 coefficients--they 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 127-meter-tall 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 -4.9t2 - 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 sense--that 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 level--1606

and normally, whatever our base level is considered--is 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 = -4.9t2 - 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 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 step--we 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 time--did 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 true--its 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 ground--until 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 solution--it 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 extraneous--we 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 right-angle 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 here--what 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 right-angle 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 "right-angle 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 is x and here is y...look, we can use the Pythagorean theorem: x2 + y2 = d2.1964

Great; so we have a way of relating these two things.1973

But if we are just frozen at 200, 2002 + 2002 = 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 x2 + y2 = d2 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 902 = (200 - 30t)2 (our red car--our north car) + (200 - 25t)2 (our east car--our 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, 902...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 t2.2116

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 + 625t2...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 625t2 + 900t2; we get 1525t2.2170

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 b2, 220002;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 one--is 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 case--if 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 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 know--I hate memorizing things, too; but it ends up coming up so often.2358

You really have to have it memorized: [-b ± √(b2 - 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 later--goodbye!2382