Hey everyone. Whenever we solved linear equations, we wanted to find a sum value for x that we could plug back into our original equation to make it true. And we want to do the same thing when solving quadratic equations. We want to find some value for x that makes our equation true. But whenever we solve linear equations, we were able to move some numbers around and then isolate x to get an answer. But what happens if I try to do the same thing with a quadratic equation? Well, looking at this x2-5x+4 over here, I might go ahead and move this 5x to the other side, maybe move my 4 as well leaving me with x2=5x-4. And then you might think, oh, we can just take the square root to get x by itself, which would leave me with x is equal to where do I go from there? What is the square root of 5x-4? So I can't actually solve a quadratic equation the same way I solved a linear equation. And more than that, there are actually often going to be 2 correct values for x when solving a quadratic equation. So how do we find those two values? Well, luckily, this something else that we need to know how to do from the previous chapter, is factoring. So let's go ahead and take a look at how we can factor to solve quadratic equations. Well, we're going to want to factor from standard form and then simply set each factor equal to 0. So if I have a quadratic equation like x2+x-6 is equal to 0 and then I go ahead and factor that into (x+3)(x-2) equals 0, I can simply take each of those individual factors, so x+3 and x-2, set them equal to 0, and then solve for x. And the reason that I can do this is because these factors are multiplied together. So x+3 times x-2 is going to equal 0 and anything times 0 is 0. So if either one of my factors, let's say that x+3 was 0, if I multiply that times anything, that would give me 0. So that might seem a little bit abstract right now. Let's go ahead and take a look at that in action. So in my example here, I want to solve this by factoring and I have x2-9x is equal to negative 20. So our first step here is going to be to write our equation in standard form. So I want to go ahead and get all my terms on the same side in descending order of power. So to do that, I just need to move my negative 20 to the other side, which I can do by adding 20 to both sides. It will cancel on the right side leaving me with x2-9x+20 is equal to 0 and we're done with step 1. Now step 2 is going to be to factor completely. Now remember, there are multiple different methods to factor, so let's go ahead and take a look at what method of factoring we should use for this equation. So we first want to look at how many terms this has. And it has this x2, negative 9x, x, and a positive twenty. So that is 3 terms, which tells me that I need to then say, does it fit a factoring formula? And looking quickly at my factoring formulas here, it doesn't seem to fit a factoring formula either. So my answer is no, which means that I then need to use the a c method to factor this quadratic. So let's go ahead and use the a c method here. So first, we want to look at a and c. So let's go ahead and make sure that we know what a, b, and c are here. So in this case, I have this kind of invisible one and that is my a, then b is negative 9 and c is 20. So if I multiply a and c together, a is just 1, so it's really just c. So a c is 20. So I want to find 2 factors that multiply to 20 and then add to b, which in this case is negative 9. So two factors that multiply to 20 add to negative 9. Since that b is negative, I know that both of my factors have to be negative, so let's think of factors of 20 that are negative. So I have negative 1 and negative 20, and then I have negative 2 and negative 10, and then I have negative 4 and negative 5. Now, negative 1 and negative 20 are going to add to negative 21, which is not negative 9. Negative 2 and negative 10 are going to add to negative 12, which is also not negative 9. But then I have negative 4 and negative 5, which do add to negative 9, so it turns out that negative 4 and negative 5 are my factors here. So once I have figured out what my factors are, I can go ahead and write this as (x-4)(x-5) is equal to 0, and step 2 is done. Now for step 3, I want to set my factors equal to 0 and then solve for x. So let's go ahead and look at each individual factor. So I have x-4. I'm setting that equal to 0, and then x-5 equals 0 as well. So let's go ahead and solve here. To solve for x with this x-4, I just need to add 4 to both sides leaving me with x is equal to 4. And then with x-5, I simply add 5 to both sides and I end up with x equals 5. And these are actually my solutions. I have completed step number 3. My solutions are x equals 4 and x equals 5. Now because we want to find 2 values for x that we could plug back into our equation that makes it true, let's go ahead and make sure that that does happen. So let's first try this x equals 4 and go ahead and plug that in. So if I plug it into my original equation, I will get 4 squared minus 9 times 4 is equal to negative 20. Now 4 squared is 16, 9 times 4 is 36. So 16 minus 36 is equal to negative 20. So negative 20 is equal to negative 20. That's definitely a true statement. I know that 4 is a value for x that makes it true. It's definitely one of my answers. Now we could do the same thing for 5 here but I'm going to leave that up to you. So that is all we need to do to solve quadratic equations by factoring. Let's go ahead and get some practice.
Table of contents
- 0. Review of Algebra4h 16m
- 1. Equations & Inequalities3h 18m
- 2. Graphs of Equations43m
- 3. Functions2h 17m
- 4. Polynomial Functions1h 44m
- 5. Rational Functions1h 23m
- 6. Exponential & Logarithmic Functions2h 28m
- 7. Systems of Equations & Matrices4h 6m
- 8. Conic Sections2h 23m
- 9. Sequences, Series, & Induction1h 19m
- 10. Combinatorics & Probability1h 45m
1. Equations & Inequalities
Intro to Quadratic Equations
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