Solving Quadratic Equations - Video Tutorials & Practice Problems
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1
concept
Introduction to Quadratic Equations
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5m
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Hey ruin. We've learned to solve a couple of different types of equations. And now we're gonna add a new equation to the mix called a quadratic equation. Now, a quadratic equation is gonna be a bit more complicated than a linear equation. And it might even seem a bit overwhelming at times. But don't worry, I'm gonna walk you through everything you need to know about quadratic equations in the next several videos starting with what a quadratic equation even is and then going into how to solve them. So let's go ahead and get started. Now, if you take a linear equation, so like two X minus six equals zero and you simply add an X squared term. So if I had a three X squared plus two X minus six equals zero, this is now a quadratic equation. Now quadratic equation is also called a polynomial of degree two because the degree or power in my equation is this two on my X squared term. So whether you hear it called a quadratic equation or a second degree polynomial, these two mean the same thing. Now you're often going to have to write quadratic equations in standard form and the standard form of a quadratic equation is AX squared plus BX plus C equals zero where all of my terms are on the same side of my equation. And they're all written in descending order of power. So my very first term A X squared has a power of two where my second term BX, I can imagine an invisible one right here. So I have a power of one and then my last term is just a constant, there is no power. So I go from 2 to 1 to 0, descending order of power. Now you're not only going to have to write quadratic equations in standard form, but you're also gonna have to be able to identify each of those coefficients A B and my constant C. So looking at my example of here, this three X squared plus two X minus six, if I asked you what a in that equation was, I would be able to say that this three, since it's the coefficient of my X squared term is A, then B is the coefficient of my X term, which in this case is this positive two, this is B and then lastly C my constant term is just the constant on the end of my equation. So in this case, it is a negative six, I always wanna make sure that I'm looking at that sign so that I can completely identify my constant or any other coefficient correctly. So let's go ahead and take a look at some other quadratic equations. So I wanna write each of these in a standard form and then identify what A B and C are. So let's take a look at our first example. Here, we have five X squared equals X minus three. Now to get this in standard form, I want all of my terms on the same side. So I'm gonna go ahead and move this X over and then my negative three over. So they're all on that left side now to get rid of the X on my right side, I need to go ahead and subtract it. Now, remember whatever you do to one side of equation you have to do to the other that has not changed. And then I need to get rid of my three by adding it. So of course, cancel on that right side, leaving me on the left side with five X squared minus X plus three equals zero because there is nothing left on that right side. Now, I of course, want these to be in descending order of power. So I have two and then my, my invisible one here and then my three doesn't have a term. So I'm good and this is my answer. So this is my quadratic equation in standard form five X squared minus X plus three equals zero. So let's go ahead and identify A B and C in this equation. So A is going to be the coefficient of the first term of my X squared term. This is also called the leading coefficient because it's the very first coefficient in my quadratic equation. So A in this equation is five, then B is the coefficient of my X term. So here I have negative X which I have an invisible one here multiplying that X so my B is negative one, then lastly my constant term is this positive three on the end here. So C here is three. And that's all for that first example. So let's go ahead and look at one more. So over here, I have negative two X squared plus five thirds is equal to zero. Now, the first thing we want to do is get all of our terms to one side and they already are here. So we're good. And then we want to check that they're in descending order of power. Well, my first term has a power of two and then my second term is just a constant, there is no power. So they actually already are written in descending order of power. And actually this is already completely in standard form. So this is my quadratic equation in standard form. Let's go ahead and identify A B and C. Now A again is the coefficient of my X squared term. So in this case, I have a negative two multiplying that X squared. So that is A B is the coefficient of my X term. But if I take a look at my equation, I actually don't have an X term at all. So B is actually going to be zero because this would be like having a plus zero X stuck in that equation, which wouldn't do anything, which is why it's not there. And then C is my constant term, which in this case is actually a fraction and it is five thirds. So it's totally fine for any of our A B and C to be a fraction or for B and C to even be zero. As long as I have that X squared term, it is still a quadratic equation. So that's all for this video. And I'll see you in the next one.
2
Problem
Problem
Write the given quadratic equation in standard form. Identify a, b, and c.
−4x2+x=8
A
a = - 4, b = 0, c = - 8
B
a = - 4, b = 1, c = 8
C
a = - 4, b = 1, c = - 8
D
a = 2, b = 1, c = 0
3
concept
Factoring
Video duration:
6m
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Hey, everyone, whenever we solved linear equations, we wanted to find some value for X that we could plug back into our original equation to make it true. And we want to do the same thing when solving quadratic equations, we want to find some value for X that makes our equation true. But whenever we solve linear equations, we were able to move some numbers around and then isolate X to get an answer. But what happens if I try to do the same thing with the quadratic equation while looking at this X squared minus five X plus four over here, I might go ahead and move this five X to the other side, maybe move my four as well. Leaving me with X squared is equal to five X minus four. And then you might think, oh, we can just take the square root to get X by itself which would leave me with X is equal to where do I go from there? What is the square root of five X minus four? So I can't actually solve a quadratic equation the same way solve a linear equation. And more than that, there are actually often going to be two correct values for X when solving a quadratic equation. So how do we find those two values? Well, luckily this something else that we need is something that we already know how to do from the previous chapter factoring. So let's go ahead and take a look at how we can factor to solve quadratic equations. Well, we're going to want to factor from standard form and then, and simply set each factor equal to zero. So if I have a quadratic equation like X squared plus X minus six is equal to zero, and then I go ahead and factor that into X plus three times X minus two equals zero. I can simply take each of those individual factors. So X plus three and X minus two, set them equal to zero and then solve for X. And the reason that I can do this is because these factors are multiplied together. So X plus three times X minus two is going to equal zero and anything times zero is zero. So if either one of my factors, let's say that X plus three was +00. If I multiply that times anything that would give me zero, so that might seem a little bit abstract right now. Let's go ahead and take a look at that in action. So in my example, here I want to solve this by factoring and I have X squared minus nine, X is equal to negative 20 So our first step here is going to be to write our equation in standard form. So I want to go ahead and get all my terms on the same side in descending order of power. So to do that, I just need to move my negative 20 to the other side, which I can do by adding 20 to both sides, it will cancel on that right side, leaving me with X squared minus nine X plus 20 is equal to zero. And we're done with step one. Now, step two is going to be to factor completely. Now remember there are multiple different methods to factor. So let's go ahead and take a look at what method of factoring we should use for this equation. So we first want to look at how many terms this has and it has this X squared negative nine X and positive 20. So that is three terms which tells me that I need to then say does it fit a factoring formula? And looking quickly at my factoring formulas here, it doesn't seem to fit a factoring formula either. So my answer is no, which means that I then need to use the AC method to factor this quadratic. So let's go ahead and use the AC method here. So first we want to look at A and C. So let's go ahead and make sure that we know what A B and C are here. So in this case, I have this kind of invisible one. And that is my A, then B is negative nine and C is 20. So if I multiply A and C together A is just one. So it's really just C so AC is 20. So I want to find two factors that multiply to 20 then add to B which in this case is negative nine. So two factors that multiply to 20 add to negative nine. Since that B is negative, I know that both of my factors have to be negative. So let's think of factors of 20 that are negative. So I have negative one and negative 20 then I have negative two and negative 10 and then I have negative four and negative five. Now negative one and negative 20 are going to add to negative 21 which is not negative nine, negative two and negative 10 are going to add to negative 12 which is also not negative nine. But then I have negative four and negative five which do add to negative nine. So it turns out that negative four and negative five are my factors here. So once I have figured out what my factors are, I can go ahead and write this as X minus four times X minus five is equal to zero and step two is done. Now for step three, I want to set my factors equal to zero and then solve for X. So let's go ahead and look at each individual factor. So I have X minus four. I'm setting that equal to zero and then X minus five equals zero as well. So let's go ahead and solve here to solve for X with this X minus four, I just need to add four to both sides. Leaving me with X is equal to four. And then with X minus five, I simply add five to both sides and I end up with X equals five. And these are actually my solutions. I have completed step number three. My solutions are X equals four and X equals five. Now, because we wanted to find two values for X that we could plug back into our equation. That makes it true. Let's go ahead and make sure that that does happen. So let's first try this X equals four and go ahead and plug that in. So if I plug it into my original equation, I will get four squared minus nine times four is equal to negative 20. Now, four squared is 16, 9 times four is 36. So 16 minus 36 is to negative 2016, minus 36 is negative 20 that's equal to negative 20. So negative 20 equals negative 20. That's definitely a true statement. I know that four is a value for X. That makes it true. It's definitely one of my answers. Now, we can do the same thing for five here, but I'm going to leave that up to you. So that is all we need to do in order to solve quadratic equations by factoring. Let's go ahead and get some practice.
4
Problem
Problem
Solve the given quadratic equation by factoring.
3x2+12x=0
A
x=3,x=4
B
x=0,x=−4
C
x=−3,x=−4
D
x=1,x=4
5
Problem
Problem
Solve the given equation by factoring.
2x2+7x+6=0
A
x=3,x=76
B
x=−2,x=0
C
x=2,x=23
D
x=−2,x=−23
6
concept
Solving Quadratic Equations by the Square Root Property
Video duration:
6m
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Hey, everyone. So we just learned how to solve quadratic equations by factoring. So if I'm given something like X squared plus X minus six equals zero, then I can just factor that into X plus three times X minus two and get my solutions from there. And that's because this equation is easily factor. But if I'm given something like X squared minus five equals zero, well, I'm not really sure what the factors would be there. And that's actually because this equation is not factor at all. Not all quadratic equations are going to be able to be solved by factoring. But that's OK because there are actually three other methods that we can use to solve quadratic equations. Now, I know that that might sound a little intimidating right now and like a lot, but I'm going to take you through each of these methods step by step and soon you'll be able to solve any quadratic equation that gets thrown at you. So let's go ahead and jump in. So we're going to start filling out this big table of all of the information that you will ever need to know about quadratic equations. And we already looked at our steps to solve a quadratic equation by factoring. But we need to look at another piece of information here, which is when we should use each method because there are multiple. So with factoring, if a quadratic equation had obvious factors or if it has no constant term. So in my standard form equation, if C is equal to zero, then factoring is going to be a good choice to solve it. But if these two things are not true, then we're going to need a new method. So here we're going to look at the square root property. Now, if those two things are not true, but I'm given something in the form of X plus A number squared is equal to a constant. So something like X plus one squared equals four or if I have no X term. So in my standard form equation B is zero and I have something like four X squared minus five equals zero. Then I'm going to go ahead and use the square root property. Now the square root property tells us that we can solve by literally just taking the square root of our quadratic equation. So let's go ahead and see that in action. So looking at my first example, here I have X plus one squared equals four. So looking at our steps, step number one is going to be to isolate our squared expression. So whatever is being squared, I want that by itself. Now, here I have X plus one being squared. So I want that by itself and it already is my squared term is by itself on that left side. So step one is done. Now step two is going to be to take both the positive and the negative square root, this is what's going to give us both of our solutions. So if I go ahead and square root my entire quadratic equation, this left side is going to cancel the square leaving me with X plus one. But then on that right side, I need to take both the positive and the negative square root. So this is going to become X plus one equals plus or minus root four. OK? So we can actually simplify this a little bit more X plus one equal to plus or minus. We know that the root four or the square root of four is just two. So this is really just plus or minus two. OK? So step two is done. Now, we can go ahead and just solve for X for step three. So here if I move my one over to the other side by just subtracting it, I, I leave X by itself and I just have X is equal to plus or minus two minus one. Now that looks a little funky the way it's written now. So if I just rearrange this and move my one over to the front, I will be left with X equals negative one plus or minus two. Now I'm actually going to be able to further simplify this if I split it into my plus and my minus and my answer. So let's go ahead and do that. So this really can become negative one plus two and negative one minus two. Those are my two solutions. Now negative one plus two is just one and then negative one minus two is negative three. So these are actually going to be my two solutions here which I can rewrite a little bit nicer as X equals one and X equals negative three. Those are my final solutions which I saw by just taking the square root. Let's take a look at one more example. So over here, I have four X squared minus five equals zero. Let's go ahead and start back at step one which is to isolate our squared expression. Now here I just have X squared. So I want to get that by itself. Now, the first thing I need to do is move by five over to the other side which I can do by adding five, it will cancel, leaving you with four, X squared is equal to five. Now, one more step to isolate that squared expression is to divide by four, leaving me with X squared is equal to 5/4. OK. So step one is done, we've isolated our squared expression. Now we can go ahead and move to step two, which is to take our positive and negative square root. So if I go ahead and square root, my quadratic equation, it will cancel on this side, leaving me with X is equal to plus or minus the square root of 5/4. Make sure you don't forget that you're taking both the positive and negative square root here. OK. So we can actually simplify this a little bit further and plus or minus root 5/4, I can just split those square roots because I know that I can further simplify the square root of four. So this is really just plus or minus root 5/2. So step two is done, we've taken our positive and negative square root and now we're left to solve for X but X is actually already by itself here. So I'm done. And these are my solutions. Now remember that you can always take your solutions and plug them back into your original equation to check and I leave that up to you. But here we're done. Our solution is just plus or minus route 5/2. I could always split that into the positive and negative solutions, but it's not gonna further simplify it here. So I'll just leave it at that. Now, you might have noticed that our answers looked a little bit different here. So on one hand, I have X equals one and X equals negative three. And over here I have the square root of 5/2. And that's actually because our solutions are not always going to be whole numbers. Sometimes they might have fractions or even radicals in them or some combination of both. And that's totally fine. Your solution doesn't have to be a whole number. So that's all you need to know about the square root property. Let's get some practice.
7
Problem
Problem
Solve the given quadratic equation using the square root property.
(x−21)2−5=0
A
x=21+5,x=21−5
B
x=25,x=−25
C
x=25,x=−25
D
x=21,x=−21
8
Problem
Problem
Solve the given quadratic equation using the square root property.
2x2−16=0
A
x=0,x=−2
B
x=42,x=−42
C
x=4,x=−4
D
x=22,x=−22
9
concept
Imaginary Roots with the Square Root Property
Video duration:
2m
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Hey, everyone, as you solve quadratic equations using the square root property, you may end up getting an imaginary or a complex root. But that's totally fine because we know how to deal with complex numbers. And we're just going to simplify it as we would any complex number. So let's look at an example here, we have four X squared plus 25 equal to zero. And I want to solve this using the square root property. So starting with step one, of course, I want to isolate my squared expression. Now my squared expression is just this X squared. So I want to go ahead and get that by itself by first moving this 25 over. And I can do that by just subtracting 25 from both sides. I'm left with four X squared is equal to negative 25 and then divide by four to get that X squared by itself. I am left with X squared is equal to negative 25/4. So step one is done, I have isolated my squared expression and now I want to take the positive and a negative square root. So taking the square root of both sides here. It will cancel on this side. I'm left with X is equal to plus or minus the square root of negative 25/4. Now, I can go ahead and simplify this further and I can separate it out into plus or minus the square root of negative 25 over the square root of four just using my radical rules. Now, I have a negative under the square root here, but it's totally fine because we've seen this before and we know how to deal with this. So let's go ahead and simplify this. Now, this square root of negative 25 is just going to become five I and then we know that the square root of four is just two. So really, this simplifies all the way into plus or minus five I over two. So we have completed step two, we've taken our positive and negative square root and X is actually already by itself. So step three is done as well. And now I just have my solution. So my solution is X is equal to plus or minus five I over two, you and I can of course, separate these into the positive and negative if I want as well. Now, whenever you're dealing with a complex answer, it's totally fine. It's fine to have the imaginary unit. And a clue that you might end up with a complex answer is if in your standard form equation A and C have the same sign. Then you're always going to end up with a complex answer. That's all there is for imaginary roots. I'll see you in the next video.
10
concept
Solving Quadratic Equations by Completing the Square
Video duration:
6m
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Hey, everyone, we just learned that whenever we have an equation in the form of X plus some number squared equals the constant, we can solve it using the square root property and simply square root both sides. But what if I have an equation that I've already determined, I can't factor. But it's also not quite in the form that I need it to be to use the square root property. Something like X squared plus six X is equal to negative seven. Well, I can actually take this equation and force it to be in that form X plus a number squared equals the constant so that I can then use the square root property something that I already know how to do. But how do we get it into that form? Well, I can actually just go ahead and add some number to both sides to make this X squared plus six X plus something factor to X plus a number squared. And this is referred to as completing the square. So I'm going to show you exactly what you need to add to both sides in order to complete the square. And then where we go from there. Let's go ahead and get started. So the first thing that we want to look at is when we're actually going to be able to use, completing the square and you can actually always complete the square. It's one of these methods that's always going to work on any quadratic equation. But there are some instances that it's going to be better to use. So if my leading coefficient A is one and B is even then completing the square is going to be a great choice. So let's go ahead and look at an example of what it actually means to complete the square. So down here, I have X squared plus six, X is equal to negative seven. Now let's check that we would even want to complete the square here. Well, my leading coefficient is definitely one and then B is six here. So that is an even number. So completing the square is going to be a great choice. So let's go ahead and take a look at step one, which is to simplify our equation to this form X squared plus BX equals C. Now, two notable things about this form are that my leading coefficient is one and C my constant is on the right side of that equation by itself. So let's go ahead and check that we have that form here. So X squared definitely already has a leading coefficient of one and my constant negative seven is already by itself on that right side. So step one is done and we can go ahead and move on to step two. Now, for step two, we're going to add A B over two squared to both sides. Now, this might seem really random. Why are we adding this B over two to both sides? But I'm going to show you exactly how it works. Let's see what happens. So here B is six. So if I take 6/2 and then square it, that's gonna give me three squared, which is just nine. So let's go ahead and add that nine to both sides. So this gives me X squared plus six, X plus nine is equal to negative seven plus nine. So we've completed step two, we've added that B over two to both sides. Step three is going to be to factor this to X plus B over two squared. So this X squared plus six X plus nine actually factors perfectly into X plus B over two squared. Here, my B was six and B over two is just 3, 6/2. So this factors into X plus three squared and that's equal to a negative seven plus nine, which is just two. So this is always how it's going to work. It is always going to factor perfectly into that B over two. And we've completed step three. Our final step here is going to be to solve using the square root property. So you might have noticed that our equation is now in the form X plus a number squared equals a constant, meaning that we can just use the square root property as we already know how. So our steps for the square root property are right here. Let's go ahead and start with step one, which is to isolate our squared expression. Now, my squared expression is actually already by itself here that X plus three squared. So step one is done and I can go ahead and move on to step two, which is to take my positive and negative square root. So square rooting both sides, I am left with X plus three is equal to plus or minus the square root of two. So step two is done as well. I've taken the positive and negative square root. Let's go ahead and solve for X. Now I can do that by moving my three over which if I subtract three from both sides, I am then just left with X on my right or on my left side there. And then I have plus or minus root two minus three. Now, we can make this look a little bit nicer by moving our negative three to the front there, leaving me with negative three plus or minus the square root of two. And I'm done. Those are my solutions negative three plus or minus the square root of two. And we've completely finished completing the square. Now, let's think about why that works for a second. So when we added this B over two squared to both sides, what we were really doing is making this into a perfect square trinomial, which is something you might remember from our factoring formulas. And it's totally OK. If you don't, you just need to know that this will always allow this to factor down into X plus B over two squared. And that's going to be equal to some constant allowing us to just use the square root property. So let's go ahead and complete the square one more time here. And here I have X squared plus eight X plus one equals zero. So let's start back at step one, which is to simplify our equation to X squared plus BX equals C. Now, here I already have that leading coefficient of one, but let's go ahead and move our constant over to the other side which we can do by subtracting it. So it will cancel, leaving me with X squared plus eight X is equal to negative one. OK? So now we can move on to step two and add B over two squared to both sides. Now, here B is eight. So if I take eight divided by two and then square it, that's going to give me four squared, which is just 16. So I'm gonna go ahead and add 16 to both sides. Now, when I add 16 to both sides, step two is completed and I can go ahead and move on to step three, which is to factor this into X plus B over two squared. Now, we already said that our B over two was this 8/2 or four. So uh this will factor into eight or X plus four squared and that's equal to negative one plus 16, which is just 15. OK. So now that we've completed step three, all we have left to do is to complete solving using the square root property. And I'm going to leave that up to you here. So that's all you need to know about completing the square. Let's go ahead and get some more practice.
11
Problem
Problem
Solve the given quadratic equation by completing the square.
x2+3x−5=0
A
x=−23,x=25
B
x=−23,x=29
C
x=2−3+29,x=2−3−29
D
x=23+29,x=23−29
12
Problem
Problem
Solve the given quadratic equation by completing the square.
3x2−6x−9=0
A
x=3,x=−1
B
x=3,x=1
C
x=2,x=3
D
x=−3,x=−4
13
concept
Quadratic Formula
Video duration:
6m
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Hey, everyone, you just finished learning three different methods of solving quadratic equations. And you might be wondering how there's anything left and why we have yet another method to learn. But the quadratic formula that we're going to talk about now is really great because it's going to work for any quadratic equation. So even if you were to forget every single other method, as long as you remember the quadratic formula, you're going to be able to solve any quadratic equation that gets thrown at you. So let's go ahead and jump in. So the quadratic formula is based on the standard form of a quadratic equation AX squared plus BX plus C. And we are going to use A B and C in order to compute our solutions. So the quadratic formula is negative B plus or minus the square root of B squared minus four times A times C all divided by two times A. Now, that formula might look a little bit complicated right now. And unfortunately, it is something that you are going to have to memorize. But that I was able to memorize the quadratic formula was by using the quadratic formula. Song now, I'm not going to sing it for you right now, but it's going to be something that you should search up on your own to commit this formula to memory. So when do we want to use the quadratic formula? Well, like I said, you can use the quadratic formula whenever you want. And some clues that you might want to use it are that you can't easily factor or you're just otherwise unsure what method to use. Let's go ahead and take a look at an example here. So I have X squared plus two, X minus three is equal to zero. Let's go ahead and take a look at our first step which is to write our equation in standard form. Now, it looks like my equation is already in standard form. All of my terms are on the same side in the in descending order of power. So I can go ahead and move on to step two. Now, step two is simply to just plug everything into my quadratic formula. So let's go ahead and do that here. I'm first going to label A B and C in my equation so that it's easy for me to just take them and plug them in. So in this case, I have an invisible one in front of that X squared and that is my A B is this positive two and then C is negative three. Make sure that you're paying attention to the science here. So plugging this in, I get negative two plus or minus the square root of two squared minus four times A, which is one times C which is a negative three all divided by two times A, which again is one. So now that we've plugged everything in, we've completed step two and all that we have left to do from here is algebra. We're just going to compute and simplify our solutions now. So let's start with what's in our um radical here and just simplify that. So this two squared is going to become a four minus four times one is just four. So really that's just four times negative three, I know that four times negative three is negative 12 and four minus negative 12, this becomes a plus. So this is really just 16. So all of that under the radical is just 16. Let's go ahead and rewrite our formula with a little more simplifying. So this is negative two plus or minus the square root of 16 because that is what we just found under the radical was all divided by two times one, which is just two. OK. So from here, I can do a couple more things. Now, I know that the square root of 16 is just four. So this becomes negative two plus or minus four all divided by two. OK. Now is when we are going to want to split our solution into the plus and the minus. So let's go ahead and do that. So this splits into negative two plus 4/2 and negative two minus 4/2. And now I can just simplify those solutions down. So negative two plus four is going to give me positive 2/2, which is just equal to one and then negative two minus four is going to give me six divided by two which is going to give me negative three. And I am done. These are my solutions. So for this quadratic equations, my solutions are X equals one and X equals negative three. And I am completely done there. I know that the quadratic formula can look a little intimidating at first, but it really just comes down to the algebra. Once you plug everything in, let's go ahead and take a look at one more example. So here I have X squared minus five X is equal to negative one. Let's go ahead and start back at step one, which is to write our equation in standard form. Now here it looks like I need to go ahead and move my negative one over. So I can do that by simply adding one to both sides. And so this will become X squared minus five X plus one is equal to zero because that one canceled on that right side. So step one is good, that's in standard form. Now we can move on to step two and just plug everything in. So again, I'm going to go ahead and label A B and C. Here I again, have an invisible one for that. A B is negative five and C is positive one. So let's go ahead and complete step two and plug everything in. So I have negative negative five. Remember to include that sign plus or minus the square root of negative five squared minus four times A which is one times C which is also one. OK. And that's all divided by two times A which again is one. OK. So step two is done. Now we're just left to simplify, just do all of that algebra. So let's first start with everything under that radical. So negative five squared is going to give me 25 minus four times one times one is just four. So this is just 25 minus four which gives me 21. So everything under the radical just simplifies 221. Let's rewrite our quadratic formula. So negative negative five is going to be positive five plus or minus the square root of 21. That is what we just found divided by two times one, which is just two. OK. So looking at this, I'm actually completely done and I can't simplify this anymore. Even if I split it, nothing would simplify because I just have that square root of 21 can't go anywhere from there. So my solutions are just five plus or minus the square root of 21/2. Now, I kind of of course, split these into their plus and minus just to show both answers separately. And I would just write five plus route 21/2 and five minus, route 21/2. Now, as you saw with some of our other methods, you're going to have different types of answers here. So you might have whole numbers, you might also have some combination of fractions and radicals and either is fine. That's all there is to the quadratic formula. Let me know if you have any questions.
14
Problem
Problem
Solve the given quadratic equation using the quadratic formula.
3x2+4x+1=0
A
x=3,x=−1
B
x=−31,x=−1
C
x=−3,x=−1
D
x=31,x=−1
15
Problem
Problem
Solve the given quadratic equation using the quadratic formula. 2x2−3x=−3
A
x=43+4i15,x=43−4i15
B
x=43+45i,x=43−45i
C
x=43+15,x=43−15
D
x=3+i15,x=3−i15
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