INSTRUCTORS Carleen Eaton Grant Fraser

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For more information, please see full course syllabus of Algebra 2

• ## Related Books

 1 answerLast reply by: Anwar AlasmariThu Jul 11, 2013 6:08 AMPost by Jose Gonzalez-Gigato on December 5, 2011At 4:15, in the section "One Rational Root", it should be -2, not -1 because it is -b. This will make the solution = -1, not - 1/2. 4 answersLast reply by: Samvel KarapetyanSun Jun 17, 2012 10:26 PMPost by Vasilios Sahinidis on December 25, 2010I thought you set a=1 and b=2, in the quadratic formula you used 1 for b. 0 answersPost by Jeffrey Petsko on September 25, 2010it needs to discuss how to slove problems like 3x squared+1=2x!

### Quadratic Formula and the Discriminant

• The quadratic formula can be used to solve anyquadratic equation.
• If you do not see an easy way to factor a quadratic equation, use the formula.
• If the discriminant is positive, the equation has 2 real roots.
• If the discriminant is 0, it has one rational root.
• If the discriminant is negative, it has two complex conjugate roots.

### Quadratic Formula and the Discriminant

Solve using the quadratic formula 3x2 + 5x = 12
• Rewrite into standard form by subtracting 12 from both sides.
• 3x2 + 5x − 12 = 0
• Find a, b and c
• a =
• b =
• c =
• a = 3
• b = 5
• c = − 12
• Plug the values into the quadratic formula x = [( − b ±√{b2 − 4ac} )/2a]
• x = [( − 5 ±√{52 − 4(3)( − 12)} )/2(3)]
• Simplify
• x = [( − 5 ±√{52 − 4(3)( − 12)} )/2(3)] = [( − 5 ±√{25 + 144} )/6] = [( − 5 ±√{169} )/6]
• Reduce square roots as much as possible
• x = [( − 5 ±√{169} )/6] = [( − 5 ±13)/6]
x = [8/6] = [4/3] and x = − 3
Solve using the quadratic formula 2x2 − 72 = 7x
• Rewrite into standard form ax2 + bx + c = 0
• 2x2 − 7x − 72 = 0
• Find a, b and c
• a =
• b =
• c =
• a = 2
• b = − 7
• c = − 72
• Plug the values into the quadratic formula x = [( − b ±√{b2 − 4ac} )/2a]
• x = [( − ( − 7) ±√{( − 7)2 − 4(2)( − 72)} )/2(2)]
• Simplify
• x = [(7 ±√{49 + 576} )/4] = [(7 ±√{625} )/4] = [(7 ±√{625} )/4]
• Reduce square roots as much as possible
• x = [(7 ±√{625} )/4] = [(7 ±25)/4]
x = [32/4] = 8 and x = − [18/4] = [9/2]
Solve using the quadratic formula 2x2 − 3 = 8x
• Rewrite into standard form ax2 + bx + c = 0
• 2x2 − 8x − 3 = 0
• Find a, b and c
• a =
• b =
• c =
• a = 2
• b = − 8
• c = − 3
• Plug the values into the quadratic formula x = [( − b ±√{b2 − 4ac} )/2a]
• x = [( − ( − 8) ±√{( − 8)2 − 4(2)( − 3)} )/2(2)]
• Simplify
• x = [(8 ±√{64 + 24} )/4] = [(8 ±√{88} )/4] =
• Reduce square roots as much as possible
• x = [(8 ±√{88} )/4] = [(8 ±√{4*22} )/4] = [(8 ±√4 *√{22} )/4] = [(8 ±2√{22} )/4]
x = [(8 + 2√{22} )/4] = [(4 + √{22} )/2] and x = [(8 − 2√{22} )/4] = [(4 − √{22} )/2]
Solve using the quadratic formula 12x2 = 17 + 8x
• Rewrite into standard form ax2 + bx + c = 0
• 12x2 − 8x − 17 = 0
• Find a, b and c
• a =
• b =
• c =
• a = 12
• b = − 8
• c = − 17
• Plug the values into the quadratic formula x = [( − b ±√{b2 − 4ac} )/2a]
• x = [( − ( − 8) ±√{( − 8)2 − 4(12)( − 17)} )/2(12)]
• Simplify
• x = [(8 ±√{64 + 816} )/24] = [(8 ±√{880} )/24] =
• Reduce square roots as much as possible
• x = [(8 ±√{880} )/24] = [(8 ±√{2*2*2*2*55} )/24] = [(8 ±4√{55} )/24] = [(8 ±4√{22} )/24] = [(4(2 ±1√{22} ))/4*6] = [(2 ±√{22} )/6]
x = [(2 + √{55} )/6] and x = [(2 − √{55} )/6]
Solve using the quadratic formula 3x2 − 11x + 8 = 0
• Find a, b and c
• a =
• b =
• c =
• a = 3
• b = − 11
• c = 8
• Plug the values into the quadratic formula x = [( − b ±√{b2 − 4ac} )/2a]
• x = [( − ( − 11) ±√{( − 11)2 − 4(3)(8)} )/2(3)]
• Simplify
• x = [(11 ±√{121 − 96} )/6] = [(11 ±√{25} )/6] =
• Reduce square roots as much as possible
• x = [(11 ±√{25} )/6] = [(11 ±5)/6]
x = [(11 + 5)/6] = [16/6] = [8/3] and x = [(11 − 5)/6] = 1
Solve using the quadratic formula 3x2 − 12x − 96 = 0
• Find a, b and c
• a =
• b =
• c =
• a = 3
• b = − 12
• c = − 96
• Plug the values into the quadratic formula x = [( − b ±√{b2 − 4ac} )/2a]
• x = [( − ( − 12) ±√{( − 12)2 − 4(3)( − 96)} )/2(3)]
• Simplify
• x = [(12 ±√{144 + 1152} )/6] = [(12 ±√{1296} )/6] =
• Reduce square roots as much as possible
• x = [(12 ±√{1296} )/6] = [(12 ±36)/6]
x = [(12 + 36)/6] = [48/6] = 8 and x = [(12 − 36)/6] = [( − 24)/6] = − 4
Solve using the quadratic formula 12x2 − 8x + 8 = 0
• Find a, b and c
• a =
• b =
• c =
• a = 12
• b = − 8
• c = 8
• Plug the values into the quadratic formula x = [( − b ±√{b2 − 4ac} )/2a]
• x = [( − ( − 8) ±√{( − 8)2 − 4(12)(8)} )/2(12)]
• Simplify
• x = [(8 ±√{64 − 384} )/24] = [(8 ±√{ − 320} )/24] = [(8 ±i√{320} )/24]
• Reduce square roots as much as possible
• x = [(8 ±i√{320} )/24] = [(8 ±i√{2*2*2*2*2*2*5} )/24] = [(8 ±8i√5 )/24] = [(8(1 ±1i√5 ))/8*3] = [(1 ±i√5 )/3]
x = [(1 + i√5 )/3] and x = [(1 − i√5 )/3]
Solve using the quadratic formula 2x2 + 4x + 8 = 0
• Find a, b and c
• a =
• b =
• c =
• a = 2
• b = 4
• c = 8
• Plug the values into the quadratic formula x = [( − b ±√{b2 − 4ac} )/2a]
• x = [( − (4) ±√{(4)2 − 4(2)(8)} )/2(2)]
• Simplify
• x = [( − 4 ±√{16 − 64} )/4] = [( − 4 ±√{ − 48} )/4] = [( − 4 ±i√{48} )/4]
• Reduce square roots as much as possible
• x = [( − 4 ±i√{48} )/4] = [( − 4 ±i√{2*2*2*2*3} )/4] = [( − 4 ±4i√3 )/4] = [(4( − 1 ±1i√3 ))/4] = − 1 ±i√3
x = − 1 − i√3 and x = − 1 + i√3
Find the discriminant and describe the nature of the solutions. − 3x2 − 3x − 12 = − 6
• Write the quadratic in standard form.
• − 3x2 − 3x − 6 = 0
• Find a, b, c
• a =
• b =
• c =
• a = − 3
• b = − 3
• c = − 6
• Calculate the discriminant D = b2 − 4ac
• D = ( − 3)2 − 4( − 3)( − 6)
• D = 9 − 72 = − 63
Since the discriminant is negative, there will be a pair of complex conjugates.
Find the discriminant and describe the nature of the solutions.
− 4x2 − 3x + 5 = 4
• Write the quadratic in standard form.
• − 4x2 − 3x + 1 = 0
• Find a, b, c
• a =
• b =
• c =
• a = − 4
• b = − 3
• c = 1
• Calculate the discriminant D = b2 − 4ac
• D = ( − 3)2 − 4( − 4)(1)
• D = 9 + 16 = 25
Since the discriminant is positive, there will be two real solutions.

*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.

### Quadratic Formula and the Discriminant

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
• Quadratic Formula 0:21
• Standard Form
• Example: Quadratic Formula
• One Rational Root 3:00
• Example: One Root
• Complex Solutions 6:16
• Complex Conjugate
• Example: Complex Solution
• Discriminant 9:42
• Positive Discriminant
• Perfect Square (Rational)
• Not Perfect Square (2 Irrational)
• Negative Discriminant
• Zero Discriminant
• Example 1: Quadratic Formula 13:50
• Example 2: Quadratic Formula 16:03
• Example 3: Quadratic Formula 19:00
• Example 4: Discriminant 21:33

### Transcription: Quadratic Formula and the Discriminant

Welcome to Educator.com.0000

Today we are going to be discussing the quadratic formula.0002

In previous lessons, we talked about solving quadratic equations through methods such as graphing and completing the square.0005

However, those methods have certain limitations that the quadratic formula does not have.0014

Let's go ahead and take a look at this.0020

Now, notice that the solutions of this quadratic equation are given by this formula, the quadratic formula.0021

However, this specifies that the equation needs to be in standard form.0030

So, make sure you have your equation in standard form before you use the quadratic formula.0033

Given that it is in standard form, and has certain values of a, b, and c, this formula will give you the solutions for this equation.0039

Now, you need to make sure that you know this well; just memorize it.0048

And then, apply it to equations in standard form.0053

Now, an example would be if I was given 2x2 + x = 2.0057

This is not in standard form; so first, put it in standard form by subtracting 2 from both sides.0062

Now, I like to write out what a, b, and c are; that way, I can just plug them in without having to worry about making errors looking back at the equation.0082

So, here a equals 1; b equals 1; and c equals -2.0089

Now, using the quadratic equation, x = -b±√(b2 [1 squared] - 4a (which is 2) c (which is -2), divided by 2a (which is 2).0094

OK, this gives me x = -1 ±...1 squared is just 1, minus...4 times 2 is 8, times -2 gives me -16; 2 times 2 is 4.0119

Here, x equals -1, plus or minus...1 minus -16 is actually positive 17, over 4.0140

Therefore, x equals (-1 ±√17)/4.0151

You can leave it like this, or you could break it out into x = (-1 + √17)/4, or x = (-1 - √17)/4.0161

So, there are two solutions; and you found them by putting the equation in standard form, and then using the quadratic formula.0170

Now, in the example I just showed you, there were actually two roots, or two solutions, to that equation.0181

However, if b2 - 4ac equals 0, the equation has only one rational root.0187

So, let's take a look at the quadratic equation again, and figure out why that is so.0194

Let's apply this to an example, x2 + 2x + 1 = 0.0207

This is already in standard form, so I don't need to worry about changing it at all.0219

Let's take a look; and I have a = 1, b = 2, and c = 1.0226

Now, notice that this b2 - 4ac is just what is here under the radical sign.0232

And this is called the discriminant; and we will talk more about that in a minute.0240

But for right now, let's just look at it and realize that it is what is under the radical sign.0243

Let's actually try to solve this out: since a equals 1, I am going to get -1 ± the square root0252

of b2 (which is 22), minus 4 times a times c, over 2 times a.0260

OK, what I end up with here is 4 - 4, which is 0.0275

So, looking at this, when I plugged in my values, I got -1 plus or minus √(4 - 4), over 2; so that is -1 plus or minus √0, over 2.0292

Well, this would come out to -1 ± 0, over 2, which just equals -1/2.0306

There is only one solution--one rational root, or one rational solution.0314

And the reason is because, if b2 - 4ac = 0, this becomes 0.0321

And the reason you have two solutions is that you take this -b/2a, and then it is plus or minus this.0327

And you would get two solutions if, for example, b2 - 4ac is 4.0335

And if I had plus or minus 4, that is going to give me something plus 2, and then something minus 2--two different things.0341

However, if I am talking about plus or minus the square root of 0, I am just going to get 0.0355

I only have one number that I am going to come out with here.0360

Therefore, if this discriminant, b2 - 4ac, is 0, you are only going to get one root, or one solution, for your quadratic equation.0363

Complex solutions may occur; and if they do, the complex solutions will actually be a pair of complex conjugates.0377

Remember what a complex conjugate pair is, such as 2 + 3i and 2 - 3i; this would be an example of complex conjugates.0388

Let's think about how this can occur, looking at the quadratic formula.0402

I have the quadratic formula; and if this ends up being a negative number, you will end up with an imaginary number.0413

And so, then you will have a real part and an imaginary part.0425

And that is how you end up with solutions that are complex conjugates.0429

For example, if I was given 2x2 + 3x + 6 = 0, then I have a = 2, b = 3, and c = 6.0433

Applying the quadratic formula to this, I am going to get -3 ± √(b2,0447

which is 32, - 4 times 2 (which is a), times 6 (which is c)), all over 2 times a, which is 2.0461

So, x equals -3, plus or minus 9; and 4 times 2 is 8, times 6; that is 48; so 9 minus 48, all over 4.0474

This equals -3 plus or minus the square root of -39.0489

You can see, right away, that this is not going to give you a real number, because it is the square root of a negative number.0499

We can simplify this further by recalling that this equals this, and also recalling that the square root of -1 equals i.0506

This simplifies, therefore, to (i√39)/4.0523

OK, now I said that the solution is a pair of complex conjugates.0535

And if we would look at this carefully, I do have a set of complex conjugates.0539

because I have x = -3, plus i√39, over 4, and x = -3 minus i√39, all of this over 4.0543

Complex solutions occur when this under the radical ends up being negative.0563

So then, you end up with an imaginary part and a real part to this number.0570

And since we have plus or minus what is under here, it ends up being a pair of complex conjugates.0574

OK, I already mentioned the discriminant; and the discriminant is that expression under the square root sign.0582

The discriminant is very helpful, because it can tell you the nature of the solutions of the equation.0589

We already saw that; but looking a little more deeply into it, if the discriminant is positive, you get two solutions, and they are both real.0596

Looking back up here at the quadratic formula, x = (-b ± √(b2 - 4ac))/2a:0622

this right here is the discriminant; so if this is positive, the square root of a positive number is a real number; so both solutions will be real.0633

Now, we can break that down further--a subset.0644

We have that, if the discriminant is positive, we get two solutions, and they are both real.0648

If the discriminant is a perfect square, then the roots are rational.0652

And that makes sense, because let's say I end up with a perfect square:0670

if the discriminant here is a perfect square 4, then I am going to end up with 2.0675

And then, I will have -b plus or minus 2, over 2a; so that makes sense.0680

Now, if the discriminant is positive, but not a perfect square, then you get two irrational roots.0685

For example, if I figured out my discriminant, b2 - 4ac, and I ended up with something like 2,0716

(it is under the square root sign--that is my discriminant), this is not a perfect square;0723

well, that is an irrational number; so I am going to get -b ± √2, over 2a.0728

So, I am going to have two solutions, and they are real numbers, but they are irrational--two irrational roots.0735

OK, if the discriminant is negative, which we just saw, then there are no real solutions.0745

Instead, you can get a pair of complex conjugates, which is what we just saw.0771

Finally, if the discriminant is 0, then you get one real solution--one real root.0776

We saw that earlier today, as well, because if this comes out to be 0, you are going to get -b/2a ±0, which is just -b/2a; there is only one solution.0793

So, if the discriminant is positive, then you have two real solutions.0804

If there is a perfect square under here, the roots are rational.0808

If it is not a perfect square, then you can get two irrational roots.0811

If the discriminant is negative, there are no real solutions, but you can get a set of complex conjugates as solutions.0818

If the discriminant is 0, there is one real root.0824

OK, let's solve this using the quadratic formula, recalling that it is x = -b ± √(b2 - 4ac), all of that over 2a.0829

Here, a is 2; b is 3; and c is -4.0846

This gives me x = -3 ± √b2 (so that is 32) - 4(2)(-4), over 2(2).0852

So, let's see what this equals: -3 plus or minus...3 times 3 is 9, minus...4 times 2 is 8, times -4 is -32, over 2 times 2, which is 4.0879

This gives me 9 minus -32; let's just say + 32, over 4.0899

So, it is -3 ± √9 + 32, over 4, which equals -3 ± √...9 + 32 is 41, over 4.0907

OK, so what I have here are two roots; I have x = (-3 + √41)/4, and x = (-3 - √41)/4.0925

And what I ended up with are two solutions; they are both real, but they are irrational,0942

because what I have in the discriminant is not a perfect square.0950

Since this is not a perfect square (it is positive, so it is real, but it is not a perfect square), I end up with irrational roots.0956

OK, again, solve using the quadratic formula.0964

The first thing I have to see is that this is not in standard form.0966

So, I am going to put it into standard form by adding 18 to both sides.0970

Now, it is in standard form, but I can actually simplify this further, because I have a common factor over here (on this left side) of 2.0982

So, I am just going to divide both sides by 2 to make it simpler.0989

And that is going to give me x2 - 6x + 9 = 0.1002

OK, now I have a = 1, b = -6, and c = 9; so by dividing, I am working with smaller numbers, which is always easier.1009

Recall the quadratic formula: OK.1018

So, let's substitute in our values here: x equals -b, so that is -(-6), plus or minus the square root of -6 squared,1029

minus 4 times a times c, all over 2 times a.1043

A negative and a negative gives me a positive; plus or minus the square root of -6 squared, which is 36, minus... 4 times 1 is 4, times 9 is 36, over 2.1053

So, this equals 6, plus or minus the square root of...36 minus 36 is 0, over 2.1071

Now, you can probably already see that the discriminant is 0 here; so I am only going to get one real root as a solution,1086

because if I say 6 plus or minus the square root of 0 (that is 0) over 2, this just comes out to 6 over 2, which is 3.1093

So, the solution here is that x equals 3.1102

So, we started out with something that was not in standard form; I put this quadratic equation into standard form.1108

And then, to make it simpler, I divided both sides by this greatest common factor that was on the left, which is 2.1115

And then, I ended up with this: x2 - 6x + 9.1124

I used that to solve using the quadratic formula, and I quickly found out that this discriminant was 0.1128

So, I ended up with one real solution, which is x = 3.1136

All right, solve using the quadratic formula again.1143

x equals -b, plus or minus the square root of b2 minus 4ac, over 2a.1148

Now, a equals 1; b equals 1--it is already in standard form; and c equals 1.1157

So, these are nice, simple numbers that I am working with.1163

x equals -b, so that is -1, plus or minus the square root of b squared (that is 1 squared)1167

minus 4 times 1 times 1 again (minus 4ac), all over 2 times a, which is 1.1176

So, x equals -1 plus or minus the square root of 1 minus 4, all over 2.1187

So, x equals -1 plus or minus the square root of 1 - 4, which is -3.1196

OK, so you can see--what happened here is that I ended up with a negative number, a negative discriminant.1203

Because it is asking me to take the square root of a negative number, my solutions are going to be a pair of complex conjugates.1209

There are no real solutions, but there are solutions.1215

Recall that I can simplify this to say -1 plus or minus √-1, times √3, over 2.1220

And I am going to recall that i equals the square root of -1.1232

Therefore, x equals -1...I am going to change this into i...so I had the square root of -3;1237

I broke that down into the square root of -1, times the square root of 3, which gives me x = -1 ± i (since this is equal to i) √3, all over 2.1253

And I can rewrite that a slightly different way, and say x = -1 + i√3, over 2, and x = -1 - i√3, all over 2.1264

So, looking at this, what I have here are solutions that are complex conjugates.1279

Here, I am asked to find the discriminant and describe the nature of the solutions.1297

So, remember that the discriminant equals b2 - 4ac.1302

This is in standard form, so here I have a = 2, b = 3, and c = 7.1309

So, my discriminant is going to be 32 - 4(2)(7); so the discriminant equals 9...4 times 2 is 8, times 7 is 56; so the discriminant equals -47.1317

Here, the discriminant is negative, so the solutions will be a pair of complex conjugates,1335

just as we saw in the last problem, because the discriminant is negative, and you are going to have1349

to take the square root of that as part of the quadratic equation.1354

And that is going to end up giving you complex conjugates.1357

OK, that concludes this lesson of Educator.com on the quadratic formula.1361

And I will see you again next time.1366