### Solving Quadratic Equations

- The principle of square roots says if we have x
^{2}= k, then x = √k or x = -√k. Note this splits it into two numbers. This method works well if the only variable is being squared. - To complete the square
- Isolate the variables on one side of the equal sign
- Divide all terms by the coefficient of the squared term
- Find the number to add to both sides of the equation. This is done by dividing the coefficient on the middle term by 2, and then squaring it.
- Factor the polynomial
- Use the principle of square roots when taking the square root of both sides
- Solve each problem for the variable separately
- The quadratic formula can be built from completing the square. Because of this it can be used to solve all quadratic equations. Make sure to property identify the coefficients used for a, b, and c.
- The discriminant can be used to determine the types of solution the quadratic equation has. Remember the discriminant is the part of the quadratic formula underneath the square root.
- Factoring is often the quickest method, but the quadratic formula will also work. Be familiar with the many ways of solving a quadratic so that you can always choose the best method for a particular quadratic equation.

### Solving Quadratic Equations

x

^{2}+ 4x + 18 = 2

- x
^{2}+ 4x = − 16 - x
^{2}+ 4x + ( [b/2] )^{2}= − 16 + ( [b/2] )^{2} - x
^{2}+ 4x + [16/4] = − 16 + [16/4] - x
^{2}+ 4x + 4 = − 16 + 4 - (x + 2)
^{2}= − 12

No Solution

y

^{2}+ 10y + 14 = 70

- y
^{2}+ 10y = 56 - y
^{2}+ 10y + ( [b/2] )^{2}= 56 - y
^{2}+ 10y + [100/4] = 56 + 25 - ( y + 5 )
^{2}= 81 - y + 5 = √{81}
- y + 5 = ±9
- y + 5 = 9y = 4
- y + 5 = − 9y = − 13

k

^{2}− 14k − 27 = 68

- k
^{2}− 14k = 95 - k
^{2}− 14k + ( [b/2] )^{2}= 95 + ( [b/2] )^{2} - k
^{2}− 14k + [196/4] = 95 + [196/4] - k
^{2}− 14k + 49 = 95 + 49 - ( k − 7 )
^{2}= 144 - k − 7 = ±12
- k − 7 = 12k = 19

r

^{2}− 8r + c

- r
^{2}− 8r + ( [b/2] )^{2} - r
^{2}− 8r + ( [( − 8)/2] )^{2} - r
^{2}− 8r + 16 - Factor:

( r + [(( − 8 ))/2] )^{2}

^{2}

g

^{2}− 5g + c

- g
^{2}− 5g + ( [b/2] )^{2} - g
^{2}− 5g + [ [(( − 5 ))/2] ]^{2} - g
^{2}− 5g + [25/4] - Factor:

( g + [(( − 5 ))/2] )^{2}

^{2}

m

^{2}+ 10m + c

- m
^{2}+ 10m + ( [b/2] )^{2} - m
^{2}+ 10m + [(10^{2})/4] - m
^{2}+ 10m + 25 - Factor:

( m + [6/2] )^{2} - ( m + [10/2] )
^{2}

^{2}

n

^{2}− 6n − 9 = 11

- n
^{2}− 6n = 20 - n
^{2}− 6n + ( [b/2] )^{2}= 20 + ( [b/2] )^{2} - n
^{2}− 6n + [6/4]^{2}= 20 + [6/4]^{2} - n
^{2}− 6n + 9 = 20 + 9 - n
^{2}− 6n + 9 = 29 - ( n + [b/2] )
^{2}= 29 - ( n + [(( − 6 ))/2] )
^{2}= 29 - ( n − 3 )
^{2}= 29 - n − 3 = ±√{29}

3x

^{2}+ 9x + 4 = 16

- 3x
^{2}+ 9x = 12 - x
^{2}+ 3x = 4 - x
^{2}+ 3x + ( [b/2] )^{2}= 4 + ( [b/2] )^{2} - x
^{2}+ 3x + [9/4] = 4 + [9/4] - x
^{2}+ 3x + [9/4] = 6[1/4] - x
^{2}+ [3/2] = ±√{6[1/4]}

v

^{2}− 12v + 3 = 39

- v
^{2}− 12v = 36 - v
^{2}− 12v + [(( 12 )^{2})/4] = 36 + [(( 12 )^{2})/4] - v
^{2}− 12v + 36 = 72 - ( v + [( − 12)/2] )
^{2}= 72 - ( v − 6 )
^{2}= 72

7w

^{2}− 56w + 16 = 37

- 7w
^{2}− 56w = 21 - w
^{2}− 8w = 3 - w
^{2}− 8w + [(8^{2})/4] = 3 + [(8^{2})/4] - w
^{2}− 8w + 16 = 19 - ( w − 4 )
^{2}= 19 - w − 4 = ±√{19}

5j

^{2}+ 6j + 1 = 0

- j = [( − 6 ±√{6
^{2}− 4( 5 )( 1 )} )/2(5)] - j = [( − 6 ±√{16} )/10]

8a

^{2}− 11a + 3 = 0

- a = [( − ( 11 ) ±√{( 11 )
^{2}− 4( 8 )( 3 )} )/2( 8 )] - a = [( − ( 11 ) ±√{25} )/16]
- a = [(11 ±5)/16]

b

^{2}+ 12b − 30 = 0

- b = [( − 12 ±√{12
^{2}− 4( 1 )( 30 )} )/2(1)]

9r

^{2}= 14r − 3

- 9r
^{2}− 14r + 3 = 0 - r = [( − ( − 14 ) ±√{( − 14 )
^{2}− 4( 9 )( 3 )} )/2( 9 )]

4s

^{2}= − 7s − 2

- 4s
^{2}+ 7s + 2 = 0 - s = [( − 7 ±√{7
^{2}− 4( 4 )( 2 )} )/2( 4 )]

20m

^{2}= 25m − 5

- 4m
^{2}= 5m − 1 - 4m
^{2}− 5m + 1 = 0 - m = [( − ( − 1 ) ±√{( − 5 )
^{2}} − 4( 4 )( 1 ))/2(4)] - m = [(1 ±√9 )/8]
- m = [(1 ±3)/8]

14y

^{2}= 49y + 21

- 2y
^{2}= 7y + 3 - 2y
^{2}− 7y − 3 = 0 - y = [( − ( − 7 ) ±√{( − 7 )
^{2}− 4( 2 )( − 3 )} )/2( 2 )]

54n

^{2}= − 81n + 18

- 6n
^{2}= 9n + 2 - 6n + 9n − 2 = 0
- n = [( − 9 ±√{9
^{2}− 4( 6 )( − 2 )} )/2( 6 )]

^{2}+ 6x + 7 = 0D = 6

^{2}− 4( 11 )( 7 )

- D = 36 − 308
- D = − 272

^{2}+ 12b − 30 = 0

- b = [( − 12 ±√{12
^{2}− 4( 1 )( 30 )} )/2( 1 )]

*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

### Solving Quadratic Equations

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
- Objectives
- Solving Quadratic Equations
- Linear Factors
- Not All Quadratics Factor Easily
- Principle of Square Roots
- Completing the Square
- Steps for Using Completing the Square
- Completing the Square Works on All Quadratic Equations
- The Quadratic Formula
- Discriminants
- Solving Quadratic Equations - Summary
- Example 1
- Example 2
- Example 3
- Example 4
- Example 5

- Intro 0:00
- Objectives 0:12
- Solving Quadratic Equations 0:29
- Linear Factors
- Not All Quadratics Factor Easily
- Principle of Square Roots
- Completing the Square
- Steps for Using Completing the Square
- Completing the Square Works on All Quadratic Equations
- The Quadratic Formula
- Discriminants
- Solving Quadratic Equations - Summary
- Example 1 11:54
- Example 2 13:03
- Example 3 16:30
- Example 4 21:29
- Example 5 25:07

### Algebra 1 Online Course

### Transcription: Solving Quadratic Equations

*Welcome back to www.educator.com.*0000

*In this lesson we are going to continue on with our solving of equations by looking at some other solving techniques.*0003

*Some of these techniques that we have seen before is of course using factoring to solve quadratic equation.*0014

*What we are going to add to that is the square root principle, how to complete a square and everyone’s favorite, the quadratic formula.*0019

*One of the first ways that you can solve the quadratic equation is break it down into its linear factors.*0032

*When you do so, you can use the principle of 0 products.*0039

*We saw that the way that it works is we treat each of the factors as a number*0042

*and we know that one or possibly both of those factors must be equal to 0.*0048

*We borrowed each of them and set them equal to 0.*0053

*Then we solved each one individually and we got our solutions.*0061

*Now of course one downside to using a process like this is that it relies on our factoring techniques.*0067

*We have to be able to factor one of those quadratics and be able to grab them and set them equal to 0.*0073

*The reason why that is so much of a problem is because not all quadratics factor easily.*0084

*In fact, some of it will factor at all for searching around.*0089

*Let us take this one like 2x ^{2} – 1 =3x.*0093

*Suppose we want to solve that quadratic equation, one of the first things we might do for trying to use that factoring technique*0097

*is we will go ahead and get everything over to one side.*0104

*Now that in itself is not too bad and then maybe we will attack this using something like reverse foil.*0110

*We only have one option to get 2x ^{2}, we must use a 2x and x.*0117

*There is only one way to get -1, we have to use 1 and -1.*0123

*The problem with this is if you go through and you check your outside and inside terms, here is what you get, -2x and 1x.*0131

*When those are combine it will give you -x, that is not the same as our middle term.*0143

*We know that factorization is not correct.*0148

*You might be saying to yourself we just made a small error.*0152

*Maybe this needs to be a -1 and the other one needs to be 1.*0155

*Well, let us try that one out and see if it turns out any better.*0163

*On the outside we have 2x, inside -1x, those combine to be 1x and unfortunately still not the same as our middle term.*0168

*There are no other possibilities for rearranging our 2 and 1, but those are the only possibilities that I could use my factoring process.*0181

*If I'm not able to factor this then how we are suppose to be able to find solutions.*0191

*After all, we still want to be able to find solutions for even quadratics like this.*0197

*For this reason alone, that we have to pick up some new methods for solving quadratics other than just the factoring process.*0202

*Let us first look at a nice one and see the method we can use on that type of quadratic.*0210

*In this type of quadratic, we only have one copy of x that is in the quadratic equation.*0219

*You do not have another x, just one copy of x ^{2}.*0225

*In these ones, the good news is you can solve directly.*0229

*This uses the principle of square roots.*0233

*What that principal says, is that if you have something like an x ^{2} equals to a number then you can split this into 2.*0237

*Just put it into x equals the square root of that number and x equals the negative square root of that number.*0247

*What you are doing in practice is like you are taking the square root of both sides of your equation.*0253

*The reason why you are getting one being positive and one being negative is the square here erases information of what the sign used to be.*0260

*Then we will go ahead and give the ± to the other side.*0274

*x could have equal to 8, but maybe it was equal to -8.*0277

*It so this was a kind of nice if you can recognize that there is only a single copy of x running around and it is being squared.*0283

*For more complicated polynomials we will have to use a very special technique.*0291

*This one is known as completing the square.*0296

*What completing square does is it puts it into a new form and we can simply apply the principles of square roots to that new form.*0299

*The process of completing the square is a little bit lengthy, so let me walk you through how the completing the square works.*0307

*The first thing that you want to do is isolate your variables to one side so get all your x’s on one side.*0315

*If it does not have a variable, go ahead and put it on the other side of the equation for now.*0321

*You want to make sure that the coefficient on your x ^{2} term is 1.*0327

*If it is anything other than 1 then go ahead and divide everything by that number.*0331

*We will be looking for a number that we can add to both sides of the equation so that it will go ahead and factor nicely.*0338

*There is a great technique for finding this number.*0345

*You take the coefficient on your x term, you divided it by 2 and you square it.*0348

*After you have added that number, then you can simply factor.*0356

*In fact, it should factor nicely at that point.*0359

*It should factor nicely.*0363

*If it does not factor nicely then double check your steps up to that one to make sure you follow them correctly.*0369

*We have gotten to that point, now we can use the principles of square roots.*0377

*This will be the square root of both sides and make sure that one of them is positive and one of them is negative.*0380

*Then we will solve directly for our variable.*0388

*This is quite a lengthy process using complete the square, but if you methodically go through all of the steps, it does get the job done.*0391

*The reason why I like completing the square so much is because it will work on any quadratic equation.*0403

*We do not have to worry if it is a special form, or if it is easily factorable, we can just use completing the square and it will work.*0410

*Of course there is one major downside to using a completing a square.*0417

*There are lots of necessary steps and it makes it very easy to make a mistake when trying to follow it.*0421

*The good news is all of these steps can be packaged up into a handy formula that we do not have quite as many steps to memorize.*0428

*This formula is known as the quadratic formula.*0436

*That is definitely something you want to know to be able to tackle many more difficult types of quadratics.*0440

*What the quadratic formula says is that if you are looking at a quadratic equation of the form ax ^{2} + bx + c = 0, its solutions are given by this.*0450

*x = b ±√(b ^{2} )- 4ac ÷2a.*0461

*It is quite a lot to keep track of but this is probably good one to memorize.*0469

*There is some good news, even though it is not the prettiest looking formula, it is built directly from completing the square.*0478

*Completing the square works on any quadratics so you may also use this on any quadratic equation.*0485

*It makes it extremely powerful for solving a lot of those quadratic equations.*0491

*Now that we have some more tools under our belt, let us take a look at using some of these new techniques.*0498

*To determine what type of solutions that these many techniques will produce, we can use what is called the discriminate to help us out.*0511

*The discriminate is actually just a part of the quadratic formula.*0520

*It is the part that sits underneath the square root.*0525

*The reason why that will tell us what type of solutions we get*0528

*is because depending on what type of numbers underneath the square root it could be real or it could be imaginary.*0533

*It could be rational or irrational.*0538

*Let us see how we can break that down.*0541

*Before looking at the part underneath the square root, that is the discriminate and say it is equal to 0.*0545

*What that tells us is that we will get exactly one real number solution.*0552

*If we are looking at the discriminate and it happens to be greater than 0, think of some positive number,*0558

*then we know we will have two different real number solutions.*0566

*This one is special because you can take it a little bit farther.*0570

*If that number is not only positive, but it is also a square number, your solutions will be rational.*0574

*If it is not a square number then it will be irrational.*0582

*And we just have one more case to worry about.*0588

*What happens if the discriminate is negative?*0590

*In that case you will have imaginary number solutions.*0593

*Think of earlier when we are dealing with a negative sign underneath the square root, these happen to be complex conjugates of each other,*0598

*but I will explain more about these in another lesson.*0606

*With so many new techniques to solve quadratics, how do you know what to use when just faced with any quadratic?*0613

*After all, some work for only certain quadratics if they are factorable and some techniques will work on any quadratic.*0621

*Here are some good tips on when you should use these.*0628

*The factoring technique that we picked up at the very beginning is one of the fastest methods you could use.*0632

*This means if you are looking at a quadratic and you can see that it is factorable*0639

*then go ahead and use that method because you will save yourself a lot of time.*0644

*If you end up struggling with it for quite a bit then maybe another technique is what should be used.*0648

*You can use the principle of square roots if you only have a single copy of x ^{2}.*0656

*This is a nice method because it is a very straightforward method that does not involve factoring and formulas.*0661

*You just solve directly.*0668

* If you are essentially stuck for too long and you know those methods worked out then you could try completing the square.*0672

*The good news is, it should be able to find your solutions, even though it does have quite a bit of steps.*0679

*In practice we do not usually use that one since the quadratic formula is built from the completing a square*0685

*and it will also work for any quadratic equation.*0690

*The reason why I like the quadratic formula it is a little bit easier to work with*0695

*and then we simply have to plugged everything into the formula and use it.*0700

*Make sure you identify your A, B and C terms and know that this one is faster than completing the square.*0705

*Let us try some of these techniques and see how it does.*0716

* We want to solve this first one using the square root principle.*0721

*To figure out why we even use that principle, notice how there is only one x and it is being squared.*0726

*There is no need to factor this one, we will just go ahead and work on getting x all by itself.*0732

*Add 5 to both sides and then we will divide both sides by 2.*0739

*And then we will take the square root of both sides.*0754

*This is where that property comes into play because I have x ^{2} equals a number.*0758

*I either have x = √5 or x = √(-5).*0764

*Let us get some more space in there.*0772

*Just solve for this one directly.*0779

*Let us try the method of completing the square with this one.*0785

*2x ^{2} + 4x – 2.*0788

*There are lots of steps involved so carefully watch how I walk through this one.*0791

*The very first thing I want to do is just start sorting things out.*0796

*I will get everything associated with a variable on one side, if it does not have a variable I will move it to the other side.*0800

*I'm adding 2 to both sides I do not have to worry about that 2 just yet.*0809

*I want to check the leading coefficient to make sure that it is a 1.*0814

*Unfortunately it looks like this one was not 1.*0819

*To make it a 1 I’m going to divide everything by 2.*0822

*2x ^{2} ÷ 2 = x^{2}, 4x ÷ 2= 2x and 2 ÷ 2 =1.*0831

*Now comes the part where I need to add something to this quadratic, so that it factors nicely.*0842

*The way we find out what this number is, is we look at the term in front of x.*0849

*I have a 2.*0853

*We are going to take that coefficient, divided by 2 and squared.*0855

*It gets a lot of questions like, should you always divide by 2 and square it?*0861

*The answer is yes it is always divided by 2 and squared.*0865

*2 ÷ 2 is 1 and 1 ^{2} is 1.*0869

*This number right over here is what I need to add to both sides of my equation.*0875

*There is all the original numbers and there is that +1 on both sides, just to keep things balanced.*0883

*The reason why we are doing that is because on the left side it will now factor nicely*0891

*You could use the reverse foil method on that left side and you will get x + 1 and x + 1.*0897

*We just get a 2.*0905

*If you do add the correct number to both sides, these factors will always be the same.*0908

*If they are not the same then go back through and check your work.*0916

*Make sure you did not make a mistake, but they will always end up being the same.*0919

*We are going to write this as x + 1 ^{2}.*0923

*That condenses things down and actually at this point this is where I can use my principle of square roots.*0930

*We are going to use it because now I can take the square root of both sides to get rid of my square.*0941

*The x + 1 = √2 , I will have x + 1= -√2.*0948

*I can solve each of these directly.*0959

*x = -1 + √2 and x= -1 -√2 .*0963

*That one does take a lot more work, especially going through the entire process of completing the square.*0974

*But notice how it did find our solutions.*0980

*It even found our solutions, despite the fact that they were irrational solutions.*0983

*Unfortunately completing the square can be a clunky process, and that is what I want to demonstrate with this next example.*0993

*2x ^{2} – 1 = 3.*0999

*Watch out even though the numbers are little bit more difficult to work with,*1002

*I will still go through all of the same steps in order to get my solution.*1005

*The very first thing I want to do is just get all of my x’s to one side.*1010

*If it does not have an x, I will move it to the other.*1014

*2x ^{2} - 3x we will add the 1 to the other side 1.*1018

*Now I need to check my x ^{2} term to see if it is 1.*1026

*It looks like this one is a 2 so I need to divide everything by 2.*1030

*That will give us x ^{2} – 3/3x = ½.*1041

*Notice how we now have a lot more fractions showing up.*1049

*We have another method wrong it is just what the numbers turns out to be.*1053

*Now that we have gotten to this point I need to figure out what number I should add to both sides of my equation*1057

*so that the left side will factor nicely.*1063

*It comes from this number right here.*1066

*We will take it, divided by 2 and take that result and square it.*1069

*Just because it is a fraction it does not mean we can not work with it,*1078

*is the same as multiplying by ½ , so I get -3/4 ^{2} which is the same as 9/16.*1082

*9/16 is the number that I should add to both sides.*1093

*Looking good, now that we have gotten to this stage the left side over here should factor nicely.*1114

*You could use method of reverse foil to help you out.*1127

*Know that your first terms will be x and two things that multiply to give you 9 would be a 3.*1132

* And two things that will multiply to give you 16 would be 4.*1140

*Both of these must be negative since our middle term is negative.*1143

*Let us see what is going on the other side.*1148

*I have ½ and I have 9/16.*1150

*What I can get those together by finding a common denominator, 8/16 + 9/16.*1154

*At this stage we have x -3/4 all of that is being squared and then all of that is equal to 17/16.*1163

*Okay so completing a square is a lot of work for us.*1175

*It gets our x’s together on the same side.*1178

*We can solve directly using the principle of square roots and taking the square root of both sides.*1183

*x - 3/4 I have the √17/16 or could be negative.*1191

*I’m going to represent those by saying it could be the + or minus of that square root.*1199

*I have one of my properties that says I can split up my square root over the top and bottom of a fraction.*1207

*The √17 or √16 or x -3/4 = ±√17÷4.*1214

*We are getting close to our actual solution, there is only one last thing I need to do.*1225

*Let us go ahead and add 3/4 to both sides.*1229

*I have x = ¾ ±√17÷4.*1241

*Since these have the same denominator I will go ahead and just write them as one large fraction.*1250

*That was quite a bit of work but notice how our answer was not a very nice answer.*1261

*It happens to be another irrational number.*1267

*We have no hope of finding irrational solutions like that using simply factoring.*1270

*Completing the square is an important process for manipulating quadratics and still being able to find solutions.*1274

*I will admit though it is quite a lengthy process, so be careful that you follow all the steps correctly.*1281

*A much faster method than completing the square is our quadratic formula.*1291

*The good news is, it will also work on any type of quadratic.*1297

*You just have to be careful to pickup the proper values and put them in their spot.*1302

*To help out, I’m going to write down the quadratic formula up in the corner here.*1305

*x= - b ±√(b ^{2} )- 4ac ÷2a.*1309

*That way we can reference it when we need to solve this one.*1323

*In order for this to work out I need it set equal to 0 first.*1327

*Let us get everything on one side.*1331

*I’m going to organize things from the largest power to the smallest power.*1334

*We will have to subtract that 7 over and now it is set equal to 0 just fine.*1339

*My A value would be 3, B value would be 2 and the C value would be -7.*1347

*I’m going to take all of those and put them into the formula.*1354

*x is equal to negative the value of B which I said was 2 + or minus the square root.*1360

*That is where we will put in B again 2 ^{2} -4 × the value of A × the value of C.*1371

*A and C all over 2 × the value of A.*1384

*You will see I just substituted the proper values into all of their spots.*1396

*We have to go through and very carefully simplify this down.*1403

*We have -2 ± 2 ^{2} = 4.*1409

*I have that -4 × 3 × -7 I have to put all of those together.*1417

*What will that give me? I have 84 ÷ 6.*1425

*- 2 ±√88÷6.*1436

*It is important to go ahead and try and simplify this as much as possible.*1445

*I have √88 in there if you look at that as 4 × 22 and be able to take care of √4 and √2 separately.*1453

*√4 would be 2 and √22 has to stay and then I can cancel out 2 from the top and the bottom.*1469

*-1±√22÷3.*1481

*You will notice how this answer was not so nice but the quadratic formula helped us find it.*1490

*You just have to be very careful that you put it in the right spot and simplify carefully.*1495

*One common mistake is watching your signs for the part underneath the square root.*1500

*One last thing to do is to use our discriminate to determine the types of solutions we will get.*1509

*No matter what method we use by the way.*1515

*The discriminate is just that part underneath the square root.*1518

*Think of b ^{2} – 4 × ac that is our discriminate.*1522

*We can use this even though the numbers might get a little large.*1528

*Here is my a, b, and c.*1532

*Let us just evaluate what this discriminate would be.*1537

*I have b ^{2}, just 70, -4 × a × c.*1541

* If I looked at 70 ^{2}, 7 × 7 is 49.*1558

* I have some 0s in there and then 4 × 25 would be 100.*1566

*100 × 49 = 49.*1571

*I'm getting for this first discriminate is that it is all equal to 0.*1576

*What is does that mean in terms of my solutions?*1582

*If I was to look at the rest of the quadratic formula in there it is telling me that this entire part is going to 0.*1584

*Either be adding 0 or subtracting 0.*1592

*I will have one real solution.*1594

*Let us try out the next one.*1605

*I have a =1, b=4, c=2.*1609

*B will be 4, -4a and c and they we will multiply all that together.*1623

*4 ^{2} is 16, 4 × 2 is 8, 16 - 8 is 8.*1633

*This number is greater than 0 but it is not a square number.*1645

*What I can say is that I will have two real solutions.*1659

*Since it is not a square number they will be irrational.*1667

*Let us do one more.*1677

*Identify a, b, and c and we will put that into our discriminate.*1680

*b ^{2} - 4 × a × c.*1692

*It is time to crunch it all down, -10 ^{2} is 100 and I have -4 × 3 × 15.*1697

*I got 12 × 15, 2 × 1, 2 + 1.*1707

*I have 100 - 180 or - 80.*1722

*In this one our discriminant is less than 0.*1729

*That is an indication that we will have two imaginary solutions.*1733

*By using the discriminant you get a good idea of the solutions that you will get in your final answer.*1747

*You can use a variety of techniques to go ahead and solve those quadratics and get your final answer.*1752

*Know that factoring is usually the fastest, but quadratic formula will work on any type of quadratics.*1758

*Thank you for watching www.educator.com.*1764

1 answer

Last reply by: Professor Eric Smith

Mon May 18, 2015 12:56 PM

Post by Lee Ross on May 18, 2015

Hi Eric,

Example 2. Just so I'm following... the second step you have written out is 2Xsqr +4X = 2. You don't combine the 2xsqr + 4x because they are not like terms? What if it was 2Xsqr + 4Xsqr = 2?