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Welcome to Algebra I: Solving Quadratic Equations Algebraically

31 minutes

Hey, guys. Welcome to Algebra I. Today's lesson is going focus on solving quadratic equations algebraically. Do you remember when you learned how to factor trinomials? That will come in handy when doing these problems. You ready? Let's go.

(Describer) Standing at a screen, she gets a stylus.

Okay. Before we start factoring quadratics, I want to jog your memory-- er, solving quadratics-- I want to jog your memory about how to factor trinomials completely.

(Describer) The screen is shown.

So, remember, a trinomial in standard form is written in this form: ax squared plus bx plus c. In this case, we have x squared plus 5x plus 6. And if you remember, when we wanted to factor a trinomial, we started out by finding factors of that c term, in this case, 6. So, we'd come off to the side. Let's get the pen. We can come off to the side and work.

(Describer) She writes.

We would start listing the factors of 6. For example, 1 times 6 will give me 6, 2 times 3 will give me 6. That's it as far as whole numbers go for factors of 6. Now we'd look back to the trinomial, find the factors of 6 that would combine to give us our middle term. We're focusing on a coefficient. In this case, we want a +5. Play around with the numbers, and I'm sure it jumps out to you that you need this pair here. If this were a +2 and this a +3, multiplying those together would give me +6. Adding those together would give me +5. I come back to my trinomial and write my factors as (x+2) and (x+3).

(Describer) ...in parentheses. ...in parentheses.

And I would have factored the trinomial completely. Now, solving quadratic equations algebraically, one of the strategies involves factoring. I wanted to make sure that you remembered how to do that process.

(Describer) She changes the screen.

What we're getting into now-- a quadratic equation. It's basically a trinomial in standard form, but it's set equal to 0. The solutions to a quadratic are the values of x that, when you substitute them into a quadratic, it gives it a value of 0. We also call the solutions to a quadratic "zeros," because if you substitute their values in there, your quadratic will have a value of 0. They're also called "roots." They have one more name I'll tell you later. For now, think about them as zeros or as roots. This is a quadratic equation.

(female) Standard form

(Describer) Standard form of a Quadratic Equation: a x-squared plus b x plus c equals zero. She changes the screen to show Example 1: x-squared plus x minus 20 equals zero.

of a quadratic equation: ax squared + bx + c = 0. Okay. Let's keep going. Let's solve one. Okay.

(female) x squared + x - 20 = 0.

(instructor) The first step when solving a quadratic equation algebraically, try factoring, right? I'm ignoring that negative temporarily, if you think about how we would factor. Let's focus on the 20.

(Describer) She writes it.

If you need more factoring practice so that these problems don't become too intense, check out that old lesson to review factoring trinomials. For now, let's keep going.

(Describer) {for now let's keep going. She draws two lines.

So, factors of 20: 1 and 20, 2 and 10, 4 and 5.

(Describer) She writes under the lines.

I know that I'm trying to get -20, so one of these factors needs to be positive, one's gonna be negative for the right pair. I'm looking for the pair that will combine to give me a +1. If you do your mental math and search through these, you'll find that it is this last pair. You'll need that 4 to be negative and that 5 positive. If you multiply those, -4 times 5, that's -20. And if you combine those, -4 plus 5, that is +1. So, now you'll go back to your quadratic, and let's rewrite the left side as a product of its factors. We're going to write it as

(x-4)(x+5) = 0.

(Describer) ...in parentheses. ...in parentheses.

How we actually figure out what the values of x are going to be for the solutions to this quadratic-- let's think about something for a second. Let's just think about properties. Remember that if we have-- Let's just pretend-- we'll come off to the side.

(Describer) She writes.

We have a times b equals 0. Just playing around for a second. If I multiply two numbers together and the answer is 0, that means that one of these numbers has to be 0, right? We use that same thinking in order to figure out the solutions to our quadratic. Because these are two factors, right? This is one factor, this is another factor. If I'm multiplying these two factors together and getting 0, that means that either this first factor is 0 or the second factor is 0. In other words, that means either x-4 is going to have to equal 0, or that means that x+5 is going to have to equal 0. One of those factors is going to have to equal 0, if when I multiply them together, the answer is 0. What we have to figure out is these values of x. We solve--I call them little mini equations or one-step equations-- We'll solve each one for x and see. If we solve that left one-- we'll add 4 to both sides, so x equals 4. And if we solve the right one, we're gonna have to subtract 5 from both sides, so x equals -5. These are our two solutions to this quadratic. If x is 4, this quadratic will have a value of 0. If x has a value of -5, it will have a value of 0. Both of these make up what we call the "solution set" to our quadratic. And let me show you how we represent that.

(Describer) She moves the screen.

Let's get this out of our way for a second. Okay. If you're asked to represent your solution using set-builder notation, this is what that means. If you stopped here, you're not necessarily wrong, depending on how you must represent your solution, but you need to be familiar with representing your solution in set-builder notation. Basically, it means "add some braces" for this kind of problem. If my solutions, or my roots, are 4 and -5, then I'll represent that as [4,-5].

(Describer) ...then a comma. ...all within braces.

I can look at that and tell, "Okay, the solutions to this quadratic are x is 4 and x is -5; one or the other." All right, let's keep solving a few more quadratics.

(female) Example 2 is x squared - 3x - 10 = 0.

(Describer) Example 2 is x-squared minus 3 x minus 10 equals zero.

Okay, so, find the solution set. If you remember the first thing we did, try to factor it, right? Let's come off to the side-- factors of 10.

(Describer) She writes.

1 and 10, 2 and 5. I need a -10, so the pair I need, one will be negative and one positive. I need them to combine and give me a -3. That means I need the second one. +2 and -5 is what I'm gonna need. I'm gonna rewrite the left side as a product of its factors. So, (x+2) and (x-5)... equals 0. Then I'm just gonna separate these and solve the two mini equations. Remember, I called them that. So, x plus 2 equals 0 and x minus 5 equals 0. So, if I solve the left one, subtract 2 from both sides... so x equals -2. Then if I solve the right one, add 5 to both sides, x equals +5. If I represent that in set-builder notation...

(Describer) She writes a left brace. ...then a comma. ...then a right brace.

[-2,5].

And you're all done. You factored it. You solved the mini equations. You got your solution set for your quadratic. Okay? Let's keep going. You should try a couple on your own. Press pause and work through these problems. To compare your answers with mine, press play.

(Describer) Title: Solve the quadratic equations. Number One: x -squared plus 2x minus 8 equals zero. Number Two: x-squared minus 7x plus 12 equals zero.

(female) Title: Solve the quadratic equations. Number one: x squared + 2x -8 = 0. Number two: x squared - 7x + 12 = 0.

(instructor) Let's see how you did. For the first one-- let's switch to the pointer tool. I'll move this.

(Describer) She uncovers the answer.

Your solution set was [-4,2]. And for the second one, the solution set was [4,3].

(Describer) She uncovers it.

If you want to see how I got these, let me show you.

(Describer) The screen just shows the first equation.

Okay. So, I started out by factoring. Switch back to my pen. Let's get the factors of 8: 1 and 8, 2 and 4. I needed a -8, so the pair I need, one will be negative, one positive. I need them to combine to give me a +2. That means I need this pair, the -2 and a +4. If I combine those, I will get a +2. If I multiple those, I will get a -8. I'll rewrite the left side as a product of its factors,

(x-2)(x+4) = 0, and I'm gonna split those up into the mini equations and solve those. So, x minus 2 equals 0, x plus 4 equals 0. Okay. Now, I'll solve. So, +2 plus 2, x equals 2, and -4 minus 4, x equals -4. So, that's how I got my solution set of [-4,2]. You may have even started to pick up on a pattern. Whatever your factors are, like how that was x-2, the solution is a +2. And that factor was x+4. The solution is a -4. That's another relationship between the factors and solutions. They're the same numbers but with opposite signs. Another pattern. Let me show you how I got that second one.

(Describer) She changes the screen.

Okay. So, begin by factoring the 12. So, 1 and 12, 2 and 6, 3 and 4. I need a +12, and I need the numbers to combine to give me a -7. Whichever pair is the right pair, both of the numbers are negative. If you run through these, you'll see it's the last one. You need a -3 and a -4. I'm gonna rewrite the left side as a product of its factors. So, (x-3)(x-4) = 0. I'm gonna split this up and solve the mini equations. So, x minus 3 equals 0, x minus 4 equals 0. Okay? Add 3 to both sides here. So, x equals 3. And add 4 to both sides here. Okay. So, x equals 4. And there is your solution set, [3,4]. Okay? All right, before we leave quadratics, there's one more thing you need to know about solving quadratic equations. If you find somebody that took Algebra I, I bet they might remember a little thing called "the quadratic formula." Before I show it, I warn you that it looks intense, but when you start running through problems using the quadratic formula, it just breaks down. It's easy to work with. So, a disclaimer first. You have to have a quadratic in standard form, and "a" cannot equal 0. Basically, you have to have a leading coefficient is what this is saying. If you have that situation, besides factoring, another method you could use to solve your quadratic equation is to use the quadratic formula, and here it is. Okay. Your solution's "x." They equal -b plus or minus the square root of b squared minus 4ac all over 2a. Okay? Take a deep breath. Let's say it one more time. So, quadratic formula: x equals -b plus or minus the square root of b squared minus 4ac all over 2a. So, in the situation where we're using the quadratic formula, we identify a, b, and c in our quadratic equation, we substitute it in this formula, and we can get our solutions that way as well. Let me show you some examples. Seeing a quadratic formula, you're probably thinking, "Why would I ever want to use that formula? It looks intense. Why don't I just factor?" Sometimes your numbers, they don't lend themselves to be factored easily. Maybe you'll have a problem that involves decimals, or it involves fractions, or it involves numbers that are quite large or quite small. Or you may be asked to show that you know how to use the quadratic formula. You can always choose whichever way you want to use. I have to make sure that you know how to use the quadratic equation. When I took Algebra I, my teacher made me memorize the quadratic formula, and I still know it now. But anytime you need to use it, write it down to help you memorize it and keep everything organized. So, before I get going, I'm going to write the quadratic formula, which I have memorized from way back in the day, but you'll have to look back and forth. Before you know it, you'll have it memorized too. So, x is gonna equal -b plus or minus the square root of b squared minus 4ac all divided by 2a. So, we're gonna use the quadratic formula to solve this equation.

(Describer) The next equation is x-squared plus 7x plus 10 equals zero.

(female) The next equation is x squared + 7x + 10 = 0. First, let's identify what a, b, and c are in our quadratic equation. Remember, when we don't see a coefficient, there's an imaginary one. Right? I know that a is 1, b is 7, and c is 10.

(Describer) She writes them all down.

Okay, half the battle you have taken care of right there by just identifying a, b, and c in your quadratic equation. Now it's all just about substituting and keeping everything organized, and then you'll be fine, okay? Let's get more space up here and start doing that. So, x = -b, and b is 7. Anytime you substitute a value in here, use parenthesis. I'm going to put -(7). Write the x so you can follow what I'm doing. Plus or minus the square root, b squared, and b is 7, so 7 squared minus 4 times a times c. Okay, so a is 1, c is 10. Make that a little longer. Come down a little bit.

(Describer) She scrolls down.

And this is all divided by 2 times a, and a was 1.

(Describer) She writes 2 times 1.

So, let's get things a little more spaced out here so it doesn't become overwhelming, or any more overwhelming than it probably already is. Let's just bring that down a bit. Keeping it spaced out will help you keep everything straight. Okay. Now it's about just cleaning up this monster that you see in front of you. Let's get the pen back. So, -b, that's basically like a -1 times 7. Don't think too much into it. It just became -7. Plus or minus. All right.

(Describer) She draws the radical sign.

Once we get that cleaned up underneath that radical, it's gonna make everything seem so easy. I already see I'm getting a little scrunched in here, so I'm just gonna move this down a little more. Okay. So, 7 squared, I know that's 49. Right? Let's get the pen back here. All right. So, 49 minus--

(Describer) She writes under the new radical.

Okay, so 4 times 1, that's 4. 4 times 10, that's 40. Okay? So, 4 times 1 is 4, 4 times 10 is 40. So, this is 49 minus 40. All right. More space. And this is all divided by 2 times 1. 2 times 1 is 2. Okay? It's already looking like less of a monster. We've got it cleaned up. So, let's keep going. Keep simplifying. A little more space here. All righty. We've still got that -7 plus or minus the square root-- So, 49 minus 40, that's just 9. Right? And this is all divided by 2. Right? Okay. Gonna keep scrolling down here so you can keep following my steps. Now we're gonna clean up that square root of 9, simplify that. So, I have -7 plus or minus-- I know the square root of 9 is 3, so plus or minus 3, all divided by 2. Now you're at the point where you split this up into two cases. You've heard me over and over again say, "Plus or minus." That literally means one time we're gonna add on the top, and one time we're gonna subtract on the top. That's how we get our solution set. So, I'm gonna write an arrow to say this implies -7 plus 3 divided by 2. We're gonna get that answer. Let's scroll down. Then we're also gonna get the subtraction case. We're gonna get -7 minus 3 divided by 2. That's what that plus or minus means. The first time, we're gonna add -7 and 3. Second time, we're gonna subtract -7 and 3. So, -7 plus 3, that's -4 divided by 2. -4 divided by 2. That's -2. Right? Now the subtraction case. So, -7 minus 3, that's -10 divided by 2. -10 divided by 2. That's -5. Okay? There we go. There are our two values for "x." That's our solution set. So, we could write our final answer, if you wrote it in set-builder notation,

[-2,-5].

And there you go. Okay? Those are the steps we would take to use the quadratic formula to solve a quadratic equation. Let's do one more together to get you more comfortable with it.

(Describer) The next equation is 3x-squared minus 11x minus 4 equals zero.

(female) The next equation is 3x squared - 11x - 4 = 0. Here we go. We're gonna use the quadratic formula to solve this quadratic equation. Okay? So first, I'm gonna write the formula. So, x equals -b plus or minus the square root of b squared minus 4ac all over 2a. All right. We've got the formula written down. Now let's come off to the side and identify a, b, and c. In this case, a is 3, b-- because it's subtract 11, we're gonna take that sign with that coefficient. So, b is -11, c is -4. Okay? All right. Half the battle's already done. We wrote down a formula, we've identified a, b, and c. Let's start substituting and cleaning things up. All right. Scroll a little bit, get a little space. I do want to keep things spaced out. Okay, so -b. So, -11 plus or minus the square root b squared, that's -11 squared minus 4... a is 3, c is -4. Okay. Now let's scroll down a bit. And this is all divided by 2a, and I believe a was 3. Yes. So, 2 times 3. Okay. We've got everything substituted in there. This is where it starts to look kind of monstrous, but we're gonna clean it up and everything will be fine. Let's get more work space here. Okay. The opposite of -11. That's how you can kind of interpret that. Opposite of -11 is +11, plus or minus the square root-- And I'm noticing things are getting crowded, so I'm just going to scoot this down a little bit. Okay. And get the pen back. All right. So, -11 squared. That's 121. I'm gonna take that negative sign with the 4 to help me do the math. So, -4 times 3, that's -12. -12 times -4, that's +48. Okay? So, plus 48. And that is all divided by 2 times 3, which is 6. Okay. Let's just keep cleaning this up. Keep scrolling. Keep breathing. Okay. All right. So, we have 11 plus or minus the square root-- 121 plus 48 is 169, and that is all divided by 6. All right. Looking better, looking better. Keep simplifying. A little more space up here. Okay. So, we have 11 plus or minus-- Okay, the square root of 169, that's 13. All divided by 6. Once we've gotten here, we're at the place where we do addition and then we do subtraction to get the two solutions. Okay. So first, 11 plus 13 divided by 6. Then we'll do 11 minus 13 divided by 6. Okay? All right, so 11 plus 13 is 24, divided by 6. 24 divided by 6, that's 4. Okay. And then the subtraction case, 11 minus 13, that's -2. -2 divided by 6. We can reduce this fraction, and it will reduce to -1/3. Okay? And there we go. We've got our solution set. So, let's write it in set-builder notation. Our solution set is [4,-1/3]. All right? Okay, I do believe it's time for you to try one. So, use the quadratic formula to solve this quadratic equation. When you're ready to compare answers with me, press play.

(Describer) The equation is 2x-squared plus 5x plus 3 equals zero.

(female) The equation is 2x squared + 5x + 3 = 0. Let's see what you got for your solution set. If you got [-1,-3/2], then you're good to go. Let me show you how I got that. Remember, we were using the quadratic formula, so we always begin by writing it down. Let's get the pen back first. I think I'll go over here this time. So, x equals -b plus or minus the square root of b squared minus 4ac all over 2a. And in this problem, our a is 2, b is 5, c is 3. Now let's just start substituting. So, -b would be -5, plus or minus the square root, b squared would be 5 squared, minus 4, a is 2, c is 3, and this is all divided by 2 times a, so 2 times 2. Okay. Just substituting should have looked something like that.

(Describer) Minus five plus or minus the square root of 5-squared minus 4 times 2 times 3, all over 2 times 2.

(female) -5 plus or minus the square root of 5 squared minus 4 times 2 times 3 all over 2 times 2. All right, let's keep going, Keep simplifying. Come down a good bit. The opposite of 5, so that's -5, plus or minus the square root. 5 squared, that's 25, minus 4 times 2, that's 8, 8 times 3, that's 24. So, 25 minus 24. All divided by 2 times 2, so that's 4. We should start feeling good when we get about right here. It's starting to look simpler. Let's keep going. Now our goal is to simplify underneath the radical there. We'll have -5 plus or minus, 25 minus 24 is 1, so that's just the square root of 1 divided by 4. Let's kind of move that down a bit. Starting to get crowded. There we go. Make this page longer. Okay. All right. Now, the square root of 1, that's just 1. Get the pen back here. So, we have -5 plus or minus 1 divided by 4. Okay. And this time I'll do the two cases just right underneath. Because we're at the point where we need to see what -5 plus 1 divided by 4 is, and we need to see what -5 minus 1 divided by 4 is. Okay. So, -5 plus 1, that's -4, divided by 4. -4 divided by 4, that's -1. -5 minus 1, that's -6. Divide it by 4, and this is a fraction you can reduce. It will reduce to -3/2. Okay? That's how we got our solution set of [-1,-3/2]. All right? Okay, good job, guys. You are now very familiar with how to solve quadratic equations algebraically by factoring or by using the quadratic formula. Hope to see you back soon for more Algebra I. Bye.

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In this program, students learn the many ways to solve quadratic equations. Some techniques include factoring, taking square roots, completing the square, and using the quadratic formula. Part of the "Welcome to Algebra I" series.

Media Details

Runtime: 31 minutes

Welcome to Algebra I
Episode 1
31 minutes
Grade Level: 7 - 12
Welcome to Algebra I
Episode 2
25 minutes
Grade Level: 7 - 12
Welcome to Algebra I
Episode 3
18 minutes
Grade Level: 7 - 12
Welcome to Algebra I
Episode 4
17 minutes
Grade Level: 7 - 12
Welcome to Algebra I
Episode 5
22 minutes
Grade Level: 7 - 12
Welcome to Algebra I
Episode 6
9 minutes
Grade Level: 7 - 12
Welcome to Algebra I
Episode 7
24 minutes
Grade Level: 7 - 12
Welcome to Algebra I
Episode 8
15 minutes
Grade Level: 7 - 12
Welcome to Algebra I
Episode 9
25 minutes
Grade Level: 7 - 12
Welcome to Algebra I
Episode 10
16 minutes
Grade Level: 7 - 12