The Quadratic Formula - Video Tutorials & Practice Problems

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Solving Quadratic Equations Using The Quadratic Formula

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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.

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Problem

Problem

Solve the given quadratic equation using the quadratic formula. $3x^2+4x+1=0$

A

$x=3,x=-1$

B

$x=-\frac13,x=-1$

C

$x=-3,x=-1$

D

$x=\frac13,x=-1$

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Problem

Problem

Solve the given quadratic equation using the quadratic formula. $2x^2-3x=-3$

Hey, everyone. So we've just finished learning about all of the different methods of solving quadratic equations. But we're going to take a quick detour here to talk about the discriminate. Now, the discriminant might feel a little bit random here and you might be wondering why you need to learn about it. But the reason I'm telling you about it at all is because you're going to be asked a very specific type of question about how many real or imaginary solutions a quadratic equation has. And you're not gonna want to have to go through fully solving it just to answer that it has one real solution. So that's where the discriminant comes in. So I'm going to show you exactly how to use the discriminate to answer this type of question really quickly. Let's go ahead and dive in. So the discriminant is just the expression underneath the radical in the quadratic formula. So it's just B squared minus four times A times C. And the sign of the discriminate is going to tell us the number and the type of solutions. So if the discriminate is positive, that means that my quadratic equation is going to have two real solutions if it's zero, that tells me that I only have one real solution. And if it's negative, that tells me that I actually have no real solutions and I have two imaginary ones. So whether it's positive zero or negative, I can see easily and quickly whether it has two real solutions, one real solution or none at all. So let's go ahead and look at a couple of different quadratic equations and identify the number and the type of solutions they have. So looking at my first one here, I have two X squared plus three X minus two equals zero. Now, if I want to calculate my discriminate here, let's go ahead and label A B and C to make it a bit easier to plug in. So A here is two, B is three, C is negative two, plugging that all into my discriminate formula. I have three squared minus four times two times negative two. Now simplifying that three squared is nine minus four times two, is eight times negative two and this becomes nine minus eight times negative two is negative 16. So this becomes nine plus 16 which gives me 25. Now this is a positive number. So that tells me that I have two real solutions. So this quadratic equation has two real solution. Let's look at another example. So I have four X squared plus X plus two equals zero. Let's go ahead and label A B and C here A is four, B is this one, I have an invisible one there and C is two. Now calculating the discriminate, I have one squared minus four times four times two, just plugging everything in. So simplifying this one squared is just one minus four times four is 16 times two and this becomes one minus 16 times two which gives me 32 and one minus 32 is going to give me negative 31. So I am left with a negative number which tells me that I have no real solutions and I have two imaginary ones. So here I have zero real and two imaginary solutions. Let's take a look at one final example here. So I have X squared minus 10, X plus 25 is equal to zero. Now, going ahead of labeling A B and C A here is one B is negative 10. Make sure to look at that sign and C is positive 25. So plugging this into my discriminate formula, I have no negative 10 squared minus four times one times 25. Now simplifying this negative 10 squared is 100 minus four times one is just four. So this is really just four times 25 which is 101 100 minus 100 is going to give me zero. So finally, here I have that my discriminate is zero, which is the last option that I have for my um discriminate and that tells me that I only have one real solution and that's all there is to the discriminate. Now, whenever you're asked this question, you can do it really quickly. Let me know if you have any questions.

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Problem

Problem

Determine the number and type of solutions of the given quadratic equation. Do not solve.

$x^2+8x+16=0$

A

2 real solutions

B

1 real solution

C

2 imaginary solutions

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Problem

Problem

Determine the number and type of solutions of the given quadratic equation. Do not solve.