Don't lose your points!

Sign up and save them.

Sign up and save them.

Or

3
Tutorials that teach
The Quadratic Formula

Take your pick:

Tutorial

Today, we're going to talk about the quadratic formula. The quadratic formula is one method that you can use for solving quadratic equations, and it's useful when you're solving quadratic equations that cannot be factored. So we're going to go through some examples of solving quadratic equations using the quadratic formula.

So for my first example, I want to solve the quadratic equation x squared plus 4x minus 5 equals 0. I can solve this equation using factoring, because I can find two integers that will multiply to give me my constant term of negative 5, and will add it to give me the value of the coefficient of my x-term, or positive 4.

So I start by listing out what will multiply to give me negative 5. That will include negative 1 and positive 5, or positive 1 and negative 5. Between these two pairs of numbers, the one that will also sum to give me a positive 4 is negative 1 and positive 5.

So I can write this equation by factoring this side of my equation with x minus 1 times x plus 5. And that is still equal to 0. Then, I can use my zero product property of multiplication, because I know that there's these two things multiplied together equal 0, that x minus 1 equals 0 or x plus 5 is equal to 0.

Solving these two equations to find my values or my solutions for x, I will add 1 to both sides here. So this will give me x is equal to 1. And here, I'll subtract 5 from both sides, which will give me x is equal to negative 5. So my solutions are x equals 1 and x equals negative 5 for this quadratic equation.

So let's look at another example of a quadratic equation. I've got x squared plus x minus 3 is equal to 0. This quadratic equation, however, cannot be solved by factoring. And that is because there are no two integers that will multiply to give me negative 3, but also add to give me positive 1. So we cannot solve this quadratic equation by factoring.

However, there is still a solution to this quadratic equation. And I know that because if I were to graph the equation y is equal to x squared plus x minus 3, I see that it crosses my x-axis at two points. And so I know that there are two solutions to this quadratic equation. It's just that the solutions are not integers, so that I cannot solve them by factoring.

So quadratic equations that cannot be solved by factoring can be solved by using something called the quadratic formula, and the quadratic formula looks like this, and will be used to find our solutions for x in our equation.

And we use the quadratic formula when our equation is written like this in standard form, and set equal to 0. And the values for a, b, and c in our formula come from the coefficients and the constant term of our quadratic equation.

We can also use the quadratic formula to determine how many solutions our equation will have. And we do that by looking at the expression underneath the radical, which we call the discriminant. b squared minus 4ac is called the discriminant.

If our discriminant is greater than 0-- strictly greater than 0-- meaning that it's positive, then we're going to have two real solutions for our quadratic equation. If our discriminant is strictly less than 0, meaning it's negative, then we're going to have no real solutions.

And that's because you cannot take the square root of a negative number. In other words, no number when it's multiplied by itself will give you a negative answer. So we can't take the square root of a negative number. And finally, if our discriminant is exactly equal to 0, then we're going to have one real solution.

So let's look at some examples of how we can use the quadratic formula to solve quadratic equations So for my first example, I've got the quadratic equation x squared plus 2x plus 3 is equal to 0. I'm going to use the quadratic formula to solve.

So I'm going to start by identifying my values for a, b, and c that I can use in my formula. And I know that the value of a will be the coefficient in front of my x squared term. My value of b will be the coefficient in front of x, and my value of C will be my constant term. And that is because my equation is already equal to 0.

My value of a will just be 1, because even though we don't write it, there's a 1 coefficient in front of x squared. So knowing my values for a ,b, and c, I can substitute that into my quadratic formula.

So I'll have x is equal to my negative b value. So negative 2 plus or minus the square root of my value of b squared. So 2 squared minus 4 times my value for a, 1, times my value for c, 3, all over 2 times my a value, which is, again, 1.

I'm going to simplify this by starting to simplify my numerator. And in my numerator, I'm going to start by simplifying the expression underneath my radical. I'll start by squaring 2. 2 squared will give me 4 minus-- then, I'm going to go ahead and multiply 4 times 1 times 3, which will give me 12.

Continuing to simplify underneath my radical, 4 minus 12 will give me a negative 8. So I have x is equal to negative 2 plus or minus the square root of negative 8 over 2 times 1.

And at this point, I can stop, because I know that if the value of my discriminant is negative-- in this case, negative 8-- then that means that I have no real solutions. So the solution to this quadratic equation is no real solution.

So for my second example, I've got the equation x squared plus 4x is equal to 3. I see this is quadratic, and I'm going to use the quadratic formula to solve. But first, I need to set this equation equal to 0.

So to do that, I'm going to subtract 3 from this side, and I'll do that also on the other side, to keep the equation balanced. On this side, I'll have 0, which is what I wanted. And on this side, I can't subtract 3 from either of these terms. So I'll simply write it at the end, making this x squared plus 4x minus 3, and then equal to 0.

So now that this equation is equal to 0, I can identify my values for a, b, and c that I'll use in my quadratic formula.

So my value for a will be the coefficient in front of my x squared term. b will be the coefficient in front of my x-term, and c will be my constant term. So I know that using my quadratic formula, a will be equal to 1, b will be equal to 4, and c will be equal to negative 3.

So now substituting these values into my formula, I will have negative 4 plus or minus the square root of 4 squared minus 4 times 1 times negative 3, all over 2 times 1. And that is going to be equal to my value or values for x.

Now I'm going to start by simplifying underneath my radical. So I have x is equal to negative plus or minus-- 4 squared will give me 16. And then, I'm going to multiply 4 times 1 times negative 3, which will give me negative 12. So I have minus negative 12, which is the same as adding a positive 12. And that's over 2 times 1.

I'll continue underneath my radical. 16 plus 12 is going to give me 28. So now, I have negative 4 plus minus the square root of 28 over 2 times 1. So taking the square root of 28 will give me approximately 5.29. So this will be approximately equal to negative 4 plus or minus 5.29, over 2 times 1.

So now, I'm going to separate this expression into two parts. One will be using the addition sign, so negative 4 plus 5.29 over-- I'm just going to simplify. 2 times 1 will give me 2. And my second solution will come from using the minus symbol. So negative 4 minus 5.29, over 2.

So now, all I need to do is simplify these two fractions. Negative 4 plus 5.29 will give me 1.29, over two, which will give me approximately 0.65 for my first solution for x.

Simplifying my second fraction, negative 4 minus 5.29 will give me a negative 9.29, over two. Dividing those two values will give me an answer of approximately negative 4.65 for my second solution for x.

So lets go over our key points from today. Prime quadratics cannot be factored because there are no two integers that will multiply to the constant term c and add to the b coefficient.

If the value of the discriminant, b squared minus 4ac, is negative, the equation has no real solution. If the value of the discriminant is non-negative, the equation has at least one real solution.

So I hope that these key points and examples helped you understand a little bit more about using the quadratic formula. Keep using your notes and keep on practicing, and soon, you'll be a pro. Thanks for watching.

Formulas to Know

- Quadratic Formula