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ELM: CSU Math Study Guide17 chapters | 147 lessons | 7 flashcard sets

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Lesson Transcript

Instructor:
*Yuanxin (Amy) Yang Alcocer*

Amy has a master's degree in secondary education and has taught math at a public charter high school.

The quadratic formula is not as complicated as it may look. You can learn to tackle this formula and succeed in using it to help you solve your quadratic problems. Watch this video to see how.

I used to hate working with the quadratic formula until I realized that it actually helped me with my problems. And sometimes, it even made my problem shorter as it gave me what I needed to know early on in the solving process. Right now, the quadratic formula might look like a beast, but think of it as a prince in disguise. It's here to help you. The **quadratic formula** allows you to solve quadratic equations in the form of *ax*^2 + *bx* + *c* = 0 and is defined as *x* = (-*b* +/- sqrt(*b*^2 - 4*ac*)) / 2*a*.

The *a*, *b*, and *c* letters come from the quadratic equation that you are solving. Let's see how this works.

The first problem we're going to solve is *x*^2 + 5*x* + 6 = 0. How do we figure out our letters? Well, we compare this to the general form of a quadratic: *ax*^2 + *bx* + *c*. A good way to do this is to write the general form above your quadratic so you can see how everything lines up. What we are looking for is the location of our *x*^2, our *x*, and our number by itself. We can then figure out what our letters stand for.

In our general form, the number attached to the *x*^2 is the letter *a*. So, looking at my quadratic, I see that there is nothing in front of the *x*^2, which means there is a 1 there, so that means my *a* = 1. Good.

Next, I look for what is front of the *x*. I see a 5, so that means my *b* = 5. The number by itself is a 6, so my *c* = 6. Now I have all my letters. If, though, you have missing terms, then that means the letter that matches that missing term is equal to zero. At this point, the quadratic formula turns into a prince to help you. To solve, all you need to do is plug in all your letters where they belong and evaluate the formula.

Let's do this.

We plug in our numbers to get *x* = (-5 +/- sqrt(5^2 - 4*1*6)) / 2*1. We begin evaluating by first calculating what is inside the square root.

*x* = (-5 +/- sqrt(25 - 24)) / 2*1

*x* = (-5 +/- sqrt (1)) / 2*1

Look at that. We have a 1 inside the square root. It is a positive number inside the square so we know we can calculate that down. If we know the square root off the top of our head, we can go ahead and write that number down. Otherwise, you can use a calculator to find the exact square root. In our case, we know that the square root of 1 is 1, so we'll write that down.

*x* = (-5 +/- 1) / 2*1

Our formula includes a +/-, so we have to split our problem into two problems, one for the plus and one for the minus. So:

*x* = (-5 + 1) / 2*1 and x = (-5 - 1) / 2*1

Now, we can finish solving these two little problems.

*x* = -4 / 2 and x = -6 / 2

So:

*x* = -2 and x = -3

And yes, we have two solutions. Most times, you will end up with two solutions like this. If they aren't whole numbers, then they'll be decimal numbers. The way you can tell that you have two solutions like this is if you get a positive number inside your square root.

There are two more cases that will give you different numbers of solutions. Let's see what happens when the square root ends up as a 0.

This time, our problem is *x*^2 + 4*x* + 4 = 0. Comparing this to our general form of *ax*^2 + *bx* + *c*, we see that our *a* = 1, our *b* = 4, and our *c* = 4. Plugging these values into our quadratic formula gives us *x* = (-4 +/- sqrt(4^2 - 4*1*4)) / 2*1. Evaluating the square root, we find that it equals zero.

*x* = (-4 +/- sqrt(16 - 16)) / 2*1

*x* = (-4 +/- sqrt(0)) / 2*1

So:

*x* = (-4 +/- 0) / 2*1

Now, we split up our problem into two and solve.

*x* = (-4 + 0) / 2*1 and *x* = (-4 - 0) / 2*1

*x* = -4 / 2 and *x* = -4 / 2

So:

*x* = -2 and *x* = -2

Hey, we have two solutions, but they're the same answer! Pretty cool, huh? So, if your square root ends up as zero, then you know that you will get the same solution twice.

There is one more special case and that is when your square root is negative inside.

When your square root is negative inside, it actually makes your problem a lot shorter. Let's see how.

Our problem is *x*^2 + 2*x* + 3 = 0. We see that our *a* = 1, our *b* = 2, and our *c* = 3. Plugging these into our quadratic formula, we get *x* = (-2 +/- sqrt(2^2 - 4*1*3)) / 2*1.

*x* = (-2 +/- sqrt(4 - 12)) / 2*1

*x* = (-2 +/- sqrt(-8)) / 2*1

Evaluating inside our square root, we see that we have a negative inside. What do we know about square roots? We know that we can't take the square root of a negative number. What does this mean for us? It means that we have to stop here, because there's no such thing as the square root of a negative number in the real world. It means that our problem has no real solutions, and we are done. We put 'no real solutions' as our answer.

In this lesson, we've learned quite a few interesting things about the quadratic formula. We know that the **quadratic formula** lets us solve problems in the general form of *ax*^2 + *bx* + *c* = 0 and is defined as *x* = (-*b* +/- sqrt(*b*^2 - 4*ac*)) / 2*a*.

We've learned that the square root part of the formula tells us how many solutions we will be getting. If it is positive, then we will have two different solutions. If it is zero, then we have two solutions, but they are the same, just repeated. And, if it is a negative inside the square root, then we have no real solutions. The quadratic formula is not such a horrible thing to work with, especially when it helps us to know if we have the right number of solutions or not.

You'll have the ability to do the following after this lesson:

- Identify the quadratic formula and explain when you can use this formula
- Explain how to determine the number of solutions based off the square root term in the quadratic formula
- Solve problems using the quadratic formula

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ELM: CSU Math Study Guide17 chapters | 147 lessons | 7 flashcard sets

- What are Polynomials, Binomials, and Quadratics? 4:39
- How to Add, Subtract and Multiply Polynomials 6:53
- Multiplying Binomials Using FOIL and the Area Method 7:26
- Multiplying Binomials Using FOIL & the Area Method: Practice Problems 5:46
- How to Factor Quadratic Equations: FOIL in Reverse 8:50
- Factoring Quadratic Equations: Polynomial Problems with a Non-1 Leading Coefficient 7:35
- How to Divide Polynomials with Long Division 8:05
- How to Use Synthetic Division to Divide Polynomials 6:51
- Dividing Polynomials with Long and Synthetic Division: Practice Problems 10:11
- How to Solve a Quadratic Equation by Factoring 7:53
- How to Solve Quadratics That Are Not in Standard Form 6:14
- How to Complete the Square 8:43
- Completing the Square Practice Problems 7:31
- How to Use the Quadratic Formula to Solve a Quadratic Equation 9:20
- Using the Quadratic Formula to Solve Equations with Literal Coefficients 5:37
- Solving Problems using the Quadratic Formula 8:32
- Go to ELM Test - Algebra: Polynomials

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