Algebra 1 (Eureka Math/EngageNY)
Course: Algebra 1 (Eureka Math/EngageNY) > Unit 4Lesson 10: Topic B: Lessons 14-15: The quadratic formula
- The quadratic formula
- Using the quadratic formula
- Worked example: quadratic formula
- Worked example: quadratic formula (example 2)
- Worked example: quadratic formula (negative coefficients)
- Quadratic formula
- Using the quadratic formula: number of solutions
- Number of solutions of quadratic equations
- Proof of the quadratic formula
The quadratic formula
The quadratic formula helps us solve any quadratic equation. First, we bring the equation to the form ax²+bx+c=0, where a, b, and c are coefficients. Then, we plug these coefficients in the formula: (-b±√(b²-4ac))/(2a) . See examples of using the formula to solve a variety of equations. Created by Sal Khan.
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- How difficult is it when you start using imaginary numbers?(23 votes)
- Don't let the term "imaginary" get in your way - there is nothing imaginary about them. They are just extensions of the real numbers, just like rational numbers (fractions) are an extension of the integers. Remember when you first started learning fractions, you encountered some different rules for adding, like the common denominator thing, as well as some other differences than the whole numbers you were used to. It seemed weird at the time, but now you are comfortable with them. Well, it is the same with imaginary numbers. They have some properties that are different from than the numbers you have been working with up to now - and that is it.
The name "imaginary number" was coined in the 17th century as a derogatory term, as such numbers were regarded by some as fictitious or useless. The term "imaginary number" now means simply a complex number with a real part equal to 0, that is, a number of the form bi.
"What's that last bit, complex number and bi" you ask?!
You'll see when you get there. Meanwhile, try this to get your feet wet:
NOTE: The Real Numbers did not have a name before Imaginary Numbers were thought of. They got called "Real" because they were not Imaginary. Really!(53 votes)
- i know how to do the quadratic formula, but my teacher gave me the problem ax squared + bx + c = 0 and she says a is not equal to zero, what are the solutions. i am not sure where to begin(16 votes)
- The solutions are just what the x values are! So you just take the quadratic equation and apply it to this. In Sal's completing the square vid, he takes the exact same equation (ax^2+bx+c = 0) and he completes the square, to end up isolating x and forming the equation into the quadratic formula. In other words, the quadratic formula is simply just ax^2+bx+c = 0 in terms of x. So the roots of ax^2+bx+c = 0 would just be the quadratic equation, which is:
(-b+-√b^2-4ac) / 2a. Hope this helped!(34 votes)
- I feel a little stupid, but how does he go from 100 to 10?4:30, thanks a lot!(20 votes)
- The square root fo 100 = 10.
Square roots reverse an exponent of 2.
Since 10^2 = 100, then square root 100 = 10.(18 votes)
- At13:35, how was he able to drop the 2 out of the equation?
- Sal skipped a couple of steps.
Factor out a GCF = 2:
[ 2 ( -6 +/- √39 )] / (-6)
The common facgtor of 2 is then cancelled with the -6 to get:
( -6 +/- √39 ) / (-3)
Hope this helps.(12 votes)
- I still do not know why this formula is important, so I'm having a hard time memorizing it.(10 votes)
- Some quadratic equations are not factorable and also would result in a mess of fractions if completing the square is used to solve them (example: 6x^2 + 7x - 8 = 0). The quadratic formula is most efficient for solving these more difficult quadratic equations.
Have a blessed, wonderful day!(11 votes)
- how to find the quadratic equation when the roots are given?(6 votes)
- Let's say that P(x) is a quadratic with roots x=a and x=b. This means that P(a)=P(b)=0. This is true if P(x) contains the factors (x - a) and (x - b), so we can write
P(x) = (x - a)(x - b).
P(a) = (a - a)(a - b) = 0(a - b) = 0.
P(b) = (b - a)(b - b) = (b - a)0 = 0.
Since P(x) = (x - a)(x - b), we can expand this and obtain
P(x) = x² - bx - ax + ab = x² - (a + b)x + ab.(16 votes)
- if the "complete the square" method always works what is the point in remembering this formula?(6 votes)
- Completing the square can get messy. You will sometimes get a lot of fractions to work thru. In those situations, the quadratic formula is often easier.(14 votes)
- I just watched the video and I can hardly remember what it is, much less how to solve it. My head is spinning on trying to figure out what it all means and how it works. Can someone else explain how it works and what to do for the problems in a different way?(5 votes)
the proof might help you understand why it works(14 votes)
- What's the main reason the Quadratic formula is used? Is there like a specific advantage for using it?(4 votes)
- Yes. You see, there are times when a quadratic may not be able to be factored (mainly a method called "completing the square"), or factoring it will produce some strange irrational results if we use the method of factoring. The quadratic formula, however, virtually gives us the same solutions, while letting us see what should be applied the square root (instead of us having to deal with the irrational values produced in an attempt to factor it).(6 votes)
- When graphing quadratic equations and also using the formula we look for x, would there be a formula for finding the y intercepts?(1 vote)
- Agreeing to Keith up there, yes, the y-intercept is very easy. When you find the x intercept, you take that y=0, right? Thereas if you are finding the y-intercept, you take x=0. If u have the quadratic equation ax^2 + bx + c = y, the y intercept = c. For example, if you had the quadratic equation x^2 + 5x - 2, you would insert x=0 so it becomes 0^2 + 5(0) - 2 = y. y=-2. Hope this wasn't confusing!(8 votes)
In this video, I'm going to expose you to what is maybe one of at least the top five most useful formulas in mathematics. And if you've seen many of my videos, you know that I'm not a big fan of memorizing things. But I will recommend you memorize it with the caveat that you also remember how to prove it, because I don't want you to just remember things and not know where they came from. But with that said, let me show you what I'm talking about: it's the quadratic formula. And as you might guess, it is to solve for the roots, or the zeroes of quadratic equations. So let's speak in very general terms and I'll show you some examples. So let's say I have an equation of the form ax squared plus bx plus c is equal to 0. You should recognize this. This is a quadratic equation where a, b and c are-- Well, a is the coefficient on the x squared term or the second degree term, b is the coefficient on the x term and then c, is, you could imagine, the coefficient on the x to the zero term, or it's the constant term. Now, given that you have a general quadratic equation like this, the quadratic formula tells us that the solutions to this equation are x is equal to negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. And I know it seems crazy and convoluted and hard for you to memorize right now, but as you get a lot more practice you'll see that it actually is a pretty reasonable formula to stick in your brain someplace. And you might say, gee, this is a wacky formula, where did it come from? And in the next video I'm going to show you where it came from. But I want you to get used to using it first. But it really just came from completing the square on this equation right there. If you complete the square here, you're actually going to get this solution and that is the quadratic formula, right there. So let's apply it to some problems. Let's start off with something that we could have factored just to verify that it's giving us the same answer. So let's say we have x squared plus 4x minus 21 is equal to 0. So in this situation-- let me do that in a different color --a is equal to 1, right? The coefficient on the x squared term is 1. b is equal to 4, the coefficient on the x-term. And then c is equal to negative 21, the constant term. And let's just plug it in the formula, so what do we get? We get x, this tells us that x is going to be equal to negative b. Negative b is negative 4-- I put the negative sign in front of that --negative b plus or minus the square root of b squared. b squared is 16, right? 4 squared is 16, minus 4 times a, which is 1, times c, which is negative 21. So we can put a 21 out there and that negative sign will cancel out just like that with that-- Since this is the first time we're doing it, let me not skip too many steps. So negative 21, just so you can see how it fit in, and then all of that over 2a. a is 1, so all of that over 2. So what does this simplify, or hopefully it simplifies? So we get x is equal to negative 4 plus or minus the square root of-- Let's see we have a negative times a negative, that's going to give us a positive. And we had 16 plus, let's see this is 6, 4 times 1 is 4 times 21 is 84. 16 plus 84 is 100. That's nice. That's a nice perfect square. All of that over 2, and so this is going to be equal to negative 4 plus or minus 10 over 2. We could just divide both of these terms by 2 right now. So this is equal to negative 4 divided by 2 is negative 2 plus or minus 10 divided by 2 is 5. So that tells us that x could be equal to negative 2 plus 5, which is 3, or x could be equal to negative 2 minus 5, which is negative 7. So the quadratic formula seems to have given us an answer for this. You can verify just by substituting back in that these do work, or you could even just try to factor this right here. You say what two numbers when you take their product, you get negative 21 and when you take their sum you get positive 4? So you'd get x plus 7 times x minus 3 is equal to negative 21. Notice 7 times negative 3 is negative 21, 7 minus 3 is positive 4. You would get x plus-- sorry it's not negative --21 is equal to 0. There should be a 0 there. So you get x plus 7 is equal to 0, or x minus 3 is equal to 0. X could be equal to negative 7 or x could be equal to 3. So it definitely gives us the same answer as factoring, so you might say, hey why bother with this crazy mess? And the reason we want to bother with this crazy mess is it'll also work for problems that are hard to factor. And let's do a couple of those, let's do some hard-to-factor problems right now. So let's scroll down to get some fresh real estate. Let's rewrite the formula again, just in case we haven't had it memorized yet. x is going to be equal to negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. I'll supply this to another problem. Let's say we have the equation 3x squared plus 6x is equal to negative 10. Well, the first thing we want to do is get it in the form where all of our terms or on the left-hand side, so let's add 10 to both sides of this equation. We get 3x squared plus the 6x plus 10 is equal to 0. And now we can use a quadratic formula. So let's apply it here. So a is equal to 3. That is a, this is b and this right here is c. So the quadratic formula tells us the solutions to this equation. The roots of this quadratic function, I guess we could call it. x is going to be equal to negative b. b is 6, so negative 6 plus or minus the square root of b squared. b is 6, so we get 6 squared minus 4 times a, which is 3 times c, which is 10. Let's stretch out the radical little bit, all of that over 2 times a, 2 times 3. So we get x is equal to negative 6 plus or minus the square root of 36 minus-- this is interesting --minus 4 times 3 times 10. So this is minus-- 4 times 3 times 10. So this is minus 120. All of that over 6. So this is interesting, you might already realize why it's interesting. What is this going to simplify to? 36 minus 120 is what? That's 84. We make this into a 10, this will become an 11, this is a 4. It is 84, so this is going to be equal to negative 6 plus or minus the square root of-- But not positive 84, that's if it's 120 minus 36. We have 36 minus 120. It's going to be negative 84 all of that 6. So you might say, gee, this is crazy. What a this silly quadratic formula you're introducing me to, Sal? It's worthless. It just gives me a square root of a negative number. It's not giving me an answer. And the reason why it's not giving you an answer, at least an answer that you might want, is because this will have no real solutions. In the future, we're going to introduce something called an imaginary number, which is a square root of a negative number, and then we can actually express this in terms of those numbers. So this actually does have solutions, but they involve imaginary numbers. So this actually has no real solutions, we're taking the square root of a negative number. So the b squared with the b squared minus 4ac, if this term right here is negative, then you're not going to have any real solutions. And let's verify that for ourselves. Let's get our graphic calculator out and let's graph this equation right here. So, let's get the graphs that y is equal to-- that's what I had there before --3x squared plus 6x plus 10. So that's the equation and we're going to see where it intersects the x-axis. Where does it equal 0? So let me graph it. Notice, this thing just comes down and then goes back up. Its vertex is sitting here above the x-axis and it's upward-opening. It never intersects the x-axis. So at no point will this expression, will this function, equal 0. At no point will y equal 0 on this graph. So once again, the quadratic formula seems to be working. Let's do one more example, you can never see enough examples here. And I want to do ones that are, you know, maybe not so obvious to factor. So let's say we get negative 3x squared plus 12x plus 1 is equal to 0. Now let's try to do it just having the quadratic formula in our brain. So the x's that satisfy this equation are going to be negative b. This is b So negative b is negative 12 plus or minus the square root of b squared, of 144, that's b squared minus 4 times a, which is negative 3 times c, which is 1, all of that over 2 times a, over 2 times negative 3. So all of that over negative 6, this is going to be equal to negative 12 plus or minus the square root of-- What is this? It's a negative times a negative so they cancel out. So I have 144 plus 12, so that is 156, right? 144 plus 12, all of that over negative 6. Now, I suspect we can simplify this 156. We could maybe bring some things out of the radical sign. So let's attempt to do that. So let's do a prime factorization of 156. Sometimes, this is the hardest part, simplifying the radical. So 156 is the same thing as 2 times 78. 78 is the same thing as 2 times what? That's 2 times 39. So the square root of 156 is equal to the square root of 2 times 2 times 39 or we could say that's the square root of 2 times 2 times the square root of 39. And this, obviously, is just going to be the square root of 4 or this is the square root of 2 times 2 is just 2. 2 square roots of 39, if I did that properly, let's see, 4 times 39. Yeah, it looks like it's right. So this up here will simplify to negative 12 plus or minus 2 times the square root of 39, all of that over negative 6. Now we can divide the numerator and the denominator maybe by 2. So this will be equal to negative 6 plus or minus the square root of 39 over negative 3. Or we could separate these two terms out. We could say this is equal to negative 6 over negative 3 plus or minus the square root of 39 over negative 3. Now, this is just a 2 right here, right? These cancel out, 6 divided by 3 is 2, so we get 2. And now notice, if this is plus and we use this minus sign, the plus will become negative and the negative will become positive. But it still doesn't matter, right? We could say minus or plus, that's the same thing as plus or minus the square root of 39 nine over 3. I think that's about as simple as we can get this answered. I want to make a very clear point of what I did that last step. I did not forget about this negative sign. I just said it doesn't matter. It's going to turn the positive into the negative; it's going to turn the negative into the positive. Let me rewrite this. So this right here can be rewritten as 2 plus the square root of 39 over negative 3 or 2 minus the square root of 39 over negative 3, right? That's what the plus or minus means, it could be this or that or both of them, really. Now in this situation, this negative 3 will turn into 2 minus the square root of 39 over 3, right? I'm just taking this negative out. Here the negative and the negative will become a positive, and you get 2 plus the square root of 39 over 3, right? A negative times a negative is a positive. So once again, you have 2 plus or minus the square of 39 over 3. 2 plus or minus the square root of 39 over 3 are solutions to this equation right there. Let verify. I'm just curious what the graph looks like. So let's just look at it. Let me clear this. Where is the clear button? So we have negative 3 three squared plus 12x plus 1 and let's graph it. Let's see where it intersects the x-axis. It goes up there and then back down again. So 2 plus or minus the square, you see-- The square root of 39 is going to be a little bit more than 6, right? Because 36 is 6 squared. So it's going be a little bit more than 6, so this is going to be a little bit more than 2. A little bit more than 6 divided by 2 is a little bit more than 2. So you're going to get one value that's a little bit more than 4 and then another value that should be a little bit less than 1. And that looks like the case, you have 1, 2, 3, 4. You have a value that's pretty close to 4, and then you have another value that is a little bit-- It looks close to 0 but maybe a little bit less than that. So anyway, hopefully you found this application of the quadratic formula helpful.