- The quadratic formula
- Understanding the 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
- Quadratic formula review
- Discriminant review
Sal determines how many solutions the equation x²+14x+49=0 has by considering its quadratic formula, and more specifically, its discriminant. Created by Sal Khan and Monterey Institute for Technology and Education.
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- So when we say we don't have "real solutions" we mean that there is no point were the parabola is touching on the x axis which is the R or real number line? And when we say we have one solution which is observable from the discriminant we say that there is only one point that touches the x axes (Real number line) and therefore it has to be its vertex?(178 votes)
- i was taught that if answer was positive it is called 2 different real numbers, if it was 0 it was called real root, and if it was negative it was imaginary numbers? is this just a different method or is this wrong...(33 votes)
- That's correct. 2 different real numbers = 2 solutions, real root = 1 solution (technically 2 but because both roots are the same there's only one solution) and imaginary numbers = no real solution (because solution is imaginary).(36 votes)
- what real life situations are there where you need the quadratic equation(26 votes)
- At3:17why is a=1?(12 votes)
- What is the meaning of having two or more answers for one equation or one expression ? and how is it applicable in the real physical world?(13 votes)
- That's a very good question. From the mathematical standpoint, two or more answers just means that there is more than one value for the variable that will solve the equation. Thinking of it graphically, if you are solving a linear equation, such as x-3 = 0, this is equivalent to looking at a line y = x-3 and seeing where it crosses the x axis (where y=0). A line can only cross another line in at most one place, so there can be at most one solution (x=3). A parabola, though, curves, so it can cross the x axis in two places. So if you have an equation like x^2 + 5x + 6 = 0, it can have two solutions.
In the real physical world, there are cases in which both these solutions will be valid, and cases where they won't, so you need to pay attention to what the solutions actually mean. For example, if you throw a ball in the air, its height over time can be described as a parabola (quadratic equation). If you are asked to calculate when it hits the ground, you will get two solutions, but both of those can't be right, because physically that makes no sense. Often in that case you might get one solution with positive time, and one solution at a negative value of time, so that one you'd throw out.
However, you might also have a problem in which there are two objects moving (maybe, two vehicles of some kind), and one is moving in a straight line and one is moving in a parabola. You might be asked where their trajectories overlap (not necessarily at the same time). Because a parabola and a line can intersect in two places, you might get two answers, and both might be correct.
I'm sure there are better real world examples! I hope this helps some, though.(28 votes)
- why does b^2-4ac < 0 no real solution when it two imaginary solutions(7 votes)
- The key word there is no "real" solution. It means the solutions are not real numbers.
It isn't saying that the is no solution at all.
Hope that clarifies.(11 votes)
- -8x^2 + 41x - 5 = (by grouping:) (-8x+1)(x-5)
(by quadratic formula:) (x-1/8)(x-5)
When I multiply the first 2 factors I get the original equation again.
Trying this with the second set of factors, i end up with opposite signs (8x^2 - 41x + 5).
What am I doing wrong ?(5 votes)
- If you expand (x-1/8)(x-5) you actually end up with x^2-(41/8)x+5/8 and not
8x^2 - 41x + 5. This makes sense because the quadratic formula will give you the roots to a quadratic equation, but remember, there are infinitely many quadratics with these roots. Think of it this way: you start with the quadratic equation: -8x^2 + 41x - 5=0 but you can do anything to rearrange this equation as long as you do it to both sides (and do not multiply by zero since that would not help at all). So if you wanted to divide both sides by 2, you would get:
-4x^2+(41/2)x-5/2 = 0 and from this equation you see that -4x^2+(41/2)x-5/2 is yet another quadratic with the same roots as -8x^2 + 41x - 5. You can keep on going and you will see that you will get different-looking quadratics with the same roots. More specifically, you get to (x-1/8)(x-5) by dividing your original quadratic -8x^2 + 41x - 5 by -8 on both sides of the quadratic equation.
(-1/8)(-8x^2 + 41x - 5) = (x-1/8)(x-5)
Hope that helps.(10 votes)
- During these kind of problems my math teacher asks whether each solution is rational and irrational, but how do I know this?(6 votes)
- When you solve a quadratic using the quadratic formula, you do a lot of simplification, including simplifying the square root. It the square root goes away because it contained a perfect square, then your answers will be rational numbers. If you still have a radical after simplifying as much as possible, then you have answers that are irrational numbers.(10 votes)
- So, Sal's solution is that x = -7, but isn't x^2 + 14x + 49 = (x + 7)(x + 7)?? Doesn't this make the solution 7?(2 votes)
- The solution (related to x-intercept, roots, and zeroes) are where y = 0. If you multiply two numbers or quantities together to get 0, one, the other, or both have to be zero. So factors of (x+7)(x+7) = 0, either x + 7 = 0 or x+7 =0, in both cases, x = -7 (subtract 7 on both sides). Factors and solutions are related, but not the same.
Also, if x = 7 and you put it back into the factors, you would get (7+7)(7+7) = 14^2 = 196 which is clearly not equal to 0. If you put -7 in, you have (-7+7)(-7+7) which is 0*0 and it does yield 0.(8 votes)
Determine the number of solutions to the quadratic equation, x squared plus 14x plus 49 is equal to 0. There's a bunch of ways we could do it. We could factor it and just figure out the values of x that satisfy it and just count them. That will be the number of solutions. We could just apply the quadratic formula. But what I want to do here is actually explore the quadratic formula, and think about how we can determine the number of solutions without even maybe necessarily finding them explicitly. So the quadratic formula tells us that if we have an equation of the form ax squared plus bx plus c is equal to 0, that the solutions are going to be-- or the solution if it exists is going to be-- negative b plus or minus the square root of b squared minus 4ac. All of that over 2a. Now the reason why this can be 2 solutions is that we have a plus or minus here. If this b squared minus 4ac is a positive number-- so let's think about this a little bit. If b squared minus 4ac is greater than 0, what's going to happen? Well, then it's a positive number. It's going to have a square root. And then when you add it to negative b you're going to get one value for the numerator, and when you subtract it from negative b you are going to get another value in the numerator. So this is going to lead to two solutions. Now what happens if b squared minus 4ac is equal to 0? If this expression under the radical is equal to 0, you're just going to have the square root of 0. So it's going to be negative b plus or minus 0. And it doesn't matter whether you add or subtract 0, you're going to get the same value. So in that situation, the actual solution of the equation is going to be negative b over 2a. There's not going to be this plus or minus, it's not going to be relevant. You're only going to have one solution. So if b squared minus 4ac is equal to 0, you only have one solution. And then what happens if b squared minus 4ac is less than 0? Well if b squared minus 4ac is less than 0, this is going to be a negative number right here and you're going to have to take the square root of a negative number. And we know, from dealing with real numbers, you can't take the square root. There is no real number squared that becomes a negative number. So in this situation there is no solutions, or no real-- when I say real I literally mean a real number-- no real solution. So let's think about it in the context of this equation right here. And just in case you're curious if whether this expression right here, b squared minus 4ac, has a name, it does. It's called the discriminant. This is the discriminant. That's that part of the quadratic equation. It determines the number of solutions we have. So if we want to figure out the number of solutions for this equation, we don't have to go through the whole quadratic equation, although it's not that much work. We just have to evaluate b squared minus 4ac. So what is b squared minus 4ac? So b is right here, it's 14. So it's 14 squared minus 4 times a, which is 1, times c, which is 49. That c, right there, times 49. What's 14 times 14? Let me do it over here. 14 times 14. 4 times 4 is 16. 4 times 1 is 4. Plus 1 is 56. Put a 0. 1 times 14 is 14. It is 6, 9, 1. It's 196. So this right here is 196. And we can ignore the 1. What's 4 times 49? So 49 times 4. 4 times 9 is 36. 4 times 4 is 16 plus 3 is 190-- or is 19, so you get 196. So this right here is 196. So b squared minus 4ac is 196 minus 196. So 196 minus 196 is equal to 0. So we're dealing with a situation where the discriminant is equal to 0. We only have one solution. And if you want, you could try to find that one solution. This whole part is going to be the square root of 0. It's just going to be 0. So the solution is going to be negative b over 2a. And negative b is-- we could just solve it. Negative b is negative 14 over 2 times a. a is just 1 over 2. So it's equal to negative 7. That's the only solution to this equation. But if you just wanted to know how many solutions, you just have to find out that b squared minus 4ac is 0. So it's only going to have one solution. And there's other ways. You could have actually factored this pretty easily into x plus 7 times x plus 7 and gotten the same result.