### Quadratic Formula and the Discriminant

- The quadratic formula can be used to solve
*any*quadratic 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

^{2}+ 5x = 12

- Rewrite into standard form by subtracting 12 from both sides.
- 3x
^{2}+ 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 ±√{b
^{2}− 4ac} )/2a] - x = [( − 5 ±√{5
^{2}− 4(3)( − 12)} )/2(3)] - Simplify
- x = [( − 5 ±√{5
^{2}− 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]

^{2}− 72 = 7x

- Rewrite into standard form ax
^{2}+ bx + c = 0 - 2x
^{2}− 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 ±√{b
^{2}− 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]

^{2}− 3 = 8x

- Rewrite into standard form ax
^{2}+ bx + c = 0 - 2x
^{2}− 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 ±√{b
^{2}− 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]

^{2}= 17 + 8x

- Rewrite into standard form ax
^{2}+ bx + c = 0 - 12x
^{2}− 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 ±√{b
^{2}− 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]

^{2}− 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 ±√{b
^{2}− 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]

^{2}− 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 ±√{b
^{2}− 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]

^{2}− 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 ±√{b
^{2}− 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]

^{2}+ 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 ±√{b
^{2}− 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

^{2}− 3x − 12 = − 6

- Write the quadratic in standard form.
- − 3x
^{2}− 3x − 6 = 0 - Find a, b, c
- a =
- b =
- c =
- a = − 3
- b = − 3
- c = − 6
- Calculate the discriminant D = b
^{2}− 4ac - D = ( − 3)
^{2}− 4( − 3)( − 6) - D = 9 − 72 = − 63

− 4x

^{2}− 3x + 5 = 4

- Write the quadratic in standard form.
- − 4x
^{2}− 3x + 1 = 0 - Find a, b, c
- a =
- b =
- c =
- a = − 4
- b = − 3
- c = 1
- Calculate the discriminant D = b
^{2}− 4ac - D = ( − 3)
^{2}− 4( − 4)(1) - D = 9 + 16 = 25

*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

### 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

### Algebra 2

### 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 2x ^{2} + 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±√(b ^{2} [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 b ^{2} - 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, x ^{2} + 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 b ^{2} - 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 root*0252

*of b ^{2} (which is 2^{2}), 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 b ^{2} - 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, b ^{2} - 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, b ^{2} - 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 2x ^{2} + 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 ± √(b ^{2},*0447

*which is 3 ^{2}, - 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 ± √(b ^{2} - 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, b ^{2} - 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 ± √(b ^{2} - 4ac), all of that over 2a.*0829

*Here, a is 2; b is 3; and c is -4.*0846

*This gives me x = -3 ± √b ^{2} (so that is 3^{2}) - 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 x ^{2} - 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: x ^{2} - 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 b ^{2} 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 b ^{2} - 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 3 ^{2} - 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 have*1349

*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

1 answer

Last reply by: Anwar Alasmari

Thu Jul 11, 2013 6:08 AM

Post by Jose Gonzalez-Gigato on December 5, 2011

At 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 answers

Last reply by: Samvel Karapetyan

Sun Jun 17, 2012 10:26 PM

Post by Vasilios Sahinidis on December 25, 2010

I thought you set a=1 and b=2, in the quadratic formula you used 1 for b.

0 answers

Post by Jeffrey Petsko on September 25, 2010

it needs to discuss how to slove problems like 3x squared+1=2x!