Solving Quadratic Equations - Online Tutor, Practice Problems & Exam Prep

Quadratic equations, or polynomials of degree 2, are expressed in standard form as a^2x + b^1x + c = 0. To solve these, methods include factoring, using the square root property, completing the square, and applying the quadratic formula: -b ± b² - 4ac / 2a. Understanding these concepts is crucial for solving various polynomial equations effectively.

Introduction to Quadratic Equations

Hey, everyone. We've learned to solve a couple of different types of equations, and now we're going to add a new equation to the mix called a quadratic equation. Now a quadratic equation is going to be a bit more complicated than a linear equation, and it might even seem a bit overwhelming at times. But don't worry, I'm going to walk you through everything you need to know about quadratic equations in the next several videos, starting with what a quadratic equation even is and then going into how to solve them. So let's go ahead and get started.

Now, if you take a linear equation, like 2x−6=0, and you simply add an x squared term, so if I had 3x2+2x−6=0, this is now a quadratic equation. A quadratic equation is also called a polynomial of degree 2 because the highest degree or power in my equation is this 2 on my x squared term. So whether you hear it called a quadratic equation or a second degree polynomial, these two mean the same thing.

You're often going to have to write quadratic equations in standard form. The standard form of a quadratic equation is ax2+bx+c=0, where all of my terms are on the same side of my equation and they're all written in descending order of power. So my first term, ax2, has a power of 2, where my second term, bx, I can imagine an invisible one right here, so I have a power of 1. And then my last term is just a constant. There is no power. So I go from 2 to 1 to 0, descending order of power.

Now, you're not only going to have to write quadratic equations in standard form but also be able to identify each of those coefficients, a, b, and my constant c. So looking at my example up here, this 3x2+2x−6, if I asked you what a in that equation was, I would be able to say that this 3, since it's the coefficient of my x squared term, is a. Then b is the coefficient of my x term, which in this case is this positive 2. This is b. And then lastly, c, my constant term, is just the constant on the end of my equation. So in this case, it is a negative 6. I always want to make sure that I'm looking at that sign so that I can completely identify my constant or any other coefficient correctly.

So let's go ahead and take a look at some other quadratic equations. Here we have 5x2 equals x−3. Now to get this in standard form, I want all of my terms on the same side. So I'm going to go ahead and move this x over and then my negative 3 over so they're all on that left side. Now to get rid of the x on my right side, I need to go ahead and subtract it. Now remember, whatever you do to one side of the equation, you have to do to the other. That has not changed. And then I need to get rid of my 3 by adding it. So we'll, of course, cancel on that right side, leaving me on the left side with 5x2−x+3=0 because there is nothing left on that right side. This is my quadratic equation in standard form. So this is my answer.

So let's go ahead and identify a, b, and c in this equation. a is going to be the coefficient of the first term of my x squared term. This is also called the leading coefficient because it's the very first coefficient in my quadratic equation. So a in this equation is 5. Then b is the coefficient of my x term. So here I have negative x, which I have an invisible one here, multiplying that x. So my b is negative 1. Then lastly, c, my constant term, is this positive 3 on the end here. So c here is 3, and that's all for that first example.

So let's go ahead and look at one more. Here, I have −2x2+53 is equal to 0. The first thing we want to do is get all of our terms to one side, and they already are here, so we're good. And then we want to check that they're in descending order of power. Well, my first term has a power of 2, and then my second term is just a constant. There is no power. So they actually already are written in descending order of power. And actually, this is already completely in standard form, so this is my quadratic equation in standard form. Let's go ahead and identify a, b, and c. Now a again is the coefficient of my x squared term. So in this case, I have a negative 2 multiplying that x squared, so that is a. B is the coefficient of my x term. But if I take a look at my equation, I actually don't have an x term at all. So b is actually going to be 0 because this would be like having a plus 0x stuck in that equation, which wouldn't do anything, which is why it's not there. And then c is my constant term, which in this case is actually a fraction, and it is 53. So it's totally fine for any of our a, b, and c to be a fraction or for b and c to even be 0. As long as I have that x squared term, it is still a quadratic equation. So that's all for this video and I'll see you in the next one.

Write the given quadratic equation in standard form. Identify a, b, and c.

$-4x^2+x=8$

• A. a = - 4, b = 0, c = - 8
• B. a = - 4, b = 1, c = 8
• C. a = - 4, b = 1, c = - 8
• D. a = 2, b = 1, c = 0

Factoring

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 minus 5x plus 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 is equal to 5x minus 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 minus 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, there is something else that we need is something that we already know how to do from the previous chapter, 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 plus x minus 6 is equal to 0 and then I go ahead and factor that into (x plus 3)(x minus 2) equals 0, I can simply take each of those individual factors, so x plus 3 and x minus 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 plus 3)(x minus 2) is going to equal 0 and anything times 0 is 0. So if either one of my factors, let's say that x plus 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 minus 9 x 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 that right side leaving me with x2 minus 9x plus 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, and a positive twenty. So that is 3 terms, which tells me that I need to then see if it fits 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]<<"response":[["this case is 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 minus 4)(x minus 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 minus 4. I'm setting that equal to 0, and then x minus 5 equals 0 as well. So let's go ahead and solve here. To solve for x with this x minus 4, I just need to add 4 to both sides leaving me with x is equal to 4. And then with x minus 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 wanted 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 4x2 minus 9 times 4 is equal to negative 20. Now 4x2 is 16, 9 times 4 is 36. So 16 minus 36 is equal to negative 20. 16 minus 36 is negative 20 and that's equal to negative 20. So negative 20 equals 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 in order to solve quadratic equations by factoring. Let's go ahead and get some practice.

Solve the given quadratic equation by factoring.

$3x^2+12x=0$

• A. $x=3,x=4$
• B. $x=0,x=-4$
• C. $x=-3,x=-4$
• D. $x=1,x=4$

Solve the given equation by factoring.

$2x^2+7x+6=0$

• A. $x=3,x=\frac67$
• B. $x=-2,x=0$
• C. $x=2,x=\frac32$
• D. $x=-2,x=-\frac32$

Solving Quadratic Equations by the Square Root Property

Hey everyone. So we just learned how to solve quadratic equations by factoring. So if I'm given something like \( x^2 + x - 6 = 0 \) then I can just factor that into \( (x + 3)(x - 2) \) and get my solutions from there. And that's because this equation is easily factorable. But if I'm given something like \( x^2 - 5 = 0 \), well, I'm not really sure what the factors would be there and that's actually because this equation is not factorable at all. Not all quadratic equations are going to be able to be solved by factoring, but that's okay because there are actually 3 other methods that we can use to solve quadratic equations. Now I know that that might sound a little intimidating right now and, like, a lot, but I'm going to take you through each of these methods step by step, and soon you'll be able to solve any quadratic equation that gets thrown at you. So let's go ahead and jump in.

So we're going to start filling out this big table of all of the information that you will ever need to know about quadratic equations. And we already looked at our steps to solve a quadratic equation by factoring, but we need to look at another piece of information here which is when we should use each method because there are multiple. So with factoring, if a quadratic equation had obvious factors or if it has no constant term, so in my standard form equation, if \( c = 0 \), then factoring is going to be a good choice to solve it. But if these two things are not true, then we're going to need a new method. So here we're going to look at the square root property.

Now if those two things are not true but I'm given something in the form of \( x + \text{a number}^2 = \text{a constant} \), so something like \( x+1^2 = 4 \). Or if I have no \( x \) term, so in my standard form equation \( b = 0 \) and I have something like \( 4x^2 - 5 = 0 \), then I'm gonna go ahead and use the square root property. Now the square root property tells us that we can solve by literally just taking the square root of our quadratic equation. So let's go ahead and see that in action.

So looking at my first example here, I have \( x + 1^2 = 4 \). So looking at our steps, step number 1 is going to be to isolate our squared expression. So whatever is being squared, I want that by itself. Now here I have \( x + 1 \) being squared, so I want that by itself and it already is. My squared term is by itself on that left side, so step 1 is done. Now step 2 is going to be to take both the positive and the negative square root. This is what's gonna give us both of our solutions. So if I go ahead and square root my entire quadratic equation, this left side is going to cancel the square, leaving me with \( x+1 \). But then on that right side, I need to take both the positive and the negative square root. So this is going to become \( x+1 = \pm \sqrt{4} \). Okay. So we can actually simplify this a little bit more. \( x + 1 = \pm 2 \). Okay so step 2 is done, now we can go ahead and just solve for \( x \) for step 3. So here if I move my one over to the other side by just subtracting it, I'm left with \( x = \pm 2 - 1 \). Now that looks a little funky the way it's written now, so if I just rearrange this and move my one over to the front, I will be left with \( x = -1 \pm 2 \). Now I'm actually going to be able to further simplify this if I split it into my plus and my minus and my answer, so let's go ahead and do that. So this really can become \(-1 + 2\) and \(-1 - 2\). Those are my two solutions. Now \(-1+2\) is just 1, and then \(-1 - 2\) is \(-3\). So these are actually gonna be my 2 solutions here, which I can rewrite a little bit nicer as \( x = 1 \) and \( x = -3 \). Those are my final solutions, which I solve by just taking the square root.

Let's take a look at one more example. So over here, I have \( 4x^2 - 5 = 0 \). Let's go ahead and start back at step 1, which is to isolate our squared expression. Now here I just have \( x^2 \), so I want to get that by itself. Now the first thing I need to do is move my 5 over to the other side, which I can do by adding 5. It will cancel leaving me with \( 4x^2 = 5 \). Now one more step to isolate that squared expression is to divide by 4, leaving me with \( x^2 = \frac{5}{4} \). Okay. So step 1 is done. We've isolated our squared expression. Now we can go ahead and move to step 2, which is to take our positive and negative square root. So if I go ahead and square root my quadratic equation, it will cancel on this side, leaving me with \( x = \pm \sqrt{\frac{5}{4}} \). Make sure you don't forget that you're taking both the positive and negative square root here. Okay, so we can actually simplify this a little bit further and \( \pm \sqrt{5}/2 \). So step 2 is done. We've taken our positive and negative square root, and now we're left to solve for \( x \). But \( x \) is actually already by itself here, so I'm done and these are my solutions. Now remember that you can always take your solutions and plug them back into your original equation to check and I will leave that up to you. But here we're done. Our solution is just \( \pm \sqrt{5}/2 \). I could always split that into the positive and negative solutions but it's not going to further simplify it here, so I'll just leave it at that.

Now you might have noticed that our answers looked a little different here. So on one hand I have \( x = 1 \) and \( x = -3 \), and over here I have \( \sqrt{5}/2 \). And that's actually because our solutions are not always going to be whole numbers. Sometimes they might have fractions or even radicals in them or some combination of both and that's totally fine, your solution doesn't have to be a whole number. So that's all you need to know about the square root property, let's get some practice.

Solve the given quadratic equation using the square root property.

$\left(x-\frac12\right)^2-5=0$

• A. $x=\frac12+\sqrt5,x=\frac12-\sqrt5$
• B. $x=\frac52,x=-\frac52$
• C. $x=\frac{\sqrt5}{2},x=-\frac{\sqrt5}{2}$
• D. $x=\frac12,x=-\frac12$

Solve the given quadratic equation using the square root property.

$2x^2-16=0$

• A. $x=0,x=-2$
• B. $x=4\sqrt2,x=-4\sqrt2$
• C. $x=4,x=-4$
• D. $x=2\sqrt2,x=-2\sqrt2$

Imaginary Roots with the Square Root Property

Hey, everyone. As you solve quadratic equations using the square root property, you may end up getting an imaginary or a complex root. But that's totally fine because we know how to deal with complex numbers, and we're just going to simplify it as we would any complex number. So let's look at an example. Here we have x2 + 4 equals 0. And I want to solve this using the square root property. So starting with step 1, of course, I want to isolate my squared expression. Now my squared expression is just this x2 so I want to go ahead and get that by itself by first moving this 25 over and I can do that by just subtracting 25 from both sides. I'm left with 4 x2 = - 25 and then divide by 4 to get that x2 by itself. I am left with x2 = - 254. So step 1 is done. I have isolated my squared expression and now I want to take the positive and negative square root. So taking the square root of both sides here it will cancel on this side. I'm left with x is equal to ± - 254. Now I can go ahead and simplify this further and I can separate it out into ± - 25 / 4 just using my radical rules. Now I have a negative under the square root here but it's totally fine because we've seen this before and we know how to deal with it. So let's go ahead and simplify this. Now this square root of negative 25 is just going to become 5i and then we know that the square root of 4 is just 2. So really, this simplifies all the way into ± 5i2. So we have completed step 2. We've taken our positive and negative square root, and x is actually already by itself so step 3 is done as well. And now I just have my solution. So my solution is x is equal to ± 5i2. And I can, of course, separate these into the positive and negative if I want as well. Now whenever you're dealing with a complex answer, it's totally fine. It's fine to have the imaginary unit. And a clue that you might end up with a complex answer is if in your imaginary unit. And a clue that you might end up with a complex answer is if in your standard form equation a and c have the same sign, then you're always going to end up with a complex answer. That's all there is for imaginary roots. I'll see you in the next video.

Solving Quadratic Equations by Completing the Square

Hey everyone. We just learned that whenever we have an equation in the form of x2+c=a constant, we can solve it using the square root property and simply take the square root of both sides. But what if I have an equation that I've already determined I can't factor but it's also not quite in the form that I need it to be to use the square root property? Something like x2+6x=-7. Well, I can actually take this equation and force it into the form x+a2=c so that I can then use the square root property, something that I already know how to do. But how do we get it into that form? Well, I can actually just go ahead and add some number to both sides to make this x2+6x+something factor to x+a2. And this is referred to as completing the square. So I'm going to show you exactly what you need to add to both sides in order to complete the square and then where we go from there. Let's go ahead and get started.

So the first thing that we want to look at is when we're actually going to be able to use completing the square, and you can actually always complete the square. It's one of these methods that's always going to work on any quadratic equation, but there are some instances that it's going to be better to use. So if my leading coefficient a is 1 and b is even, then completing the square is going to be a great choice. So let's go ahead and look at an example of what it actually means to complete the square.

So down here I have x2+6x=-7. Now let's check that we would even want to complete the square here. Well, my leading coefficient is definitely 1, and then b is 6 here, so that is an even number. So completing the square is going to be a great choice. So let's go ahead and take a look at step 1, which is to simplify our equation to this form, x2+bx=c. Now two notable things about this form are that my leading coefficient is 1, and c, my constant, is on the right side of that equation by itself.

So let's go ahead and check that we have that form here. So x2 definitely already has a leading coefficient of 1, and my 2, we're going to add (b/2)2 to both sides. Now this might seem really random, why are we adding this b/2 to both sides? But I'm going to show you exactly how it works. Let's see what happens. So here b is 6, so if I take 6 divided by 2 and then square it, that's going to give me 32 which is just 9. So let's go ahead and add that 9 to both sides. So this gives me x2+6x+9=-7+9. So we've completed step 2. We've added that b/2 to both sides. Step 3 is going to be to factor this to x+b/22. So this x2+6x+9 actually factors perfectly into x+b/22. Here my b was 6 and b/2 is just 3, 6 divided by 2, so this factors into x+32=2. So this is always how it's going to work. It is always going to factor perfectly into that b/2 and we've completed step 3. Our final step here is going to be to solve using the square root property.

So you might have noticed that our equation is now in the form x+a2=c, meaning that we can just use the square root property as we already know how. So our steps for the square root property are right here. Let's go ahead and start with step 1, which is to isolate our squared expression. Now, my squared expression is actually already by itself here, that x+32. So step 1 is done, and I can go ahead and move on to step 2, which is to take the positive and negative square root. So, square-rooting both sides, I am left with x+3=±2. So step 2 is done as well. I've taken the positive and negative square root. Let's go ahead and solve for x. Now I can do that by moving my 3 over, which if I subtract 3 from both sides, I am then just left with x on my left side there. And then I have ±2-3. Now we can make this look a little nicer by moving our negative 3 to the front there, leaving me with -3±2, and I'm done. Those are my solutions. -3±2 and we've completely finished completing the square.

Now let's think about why that works for a second. So when we added this b/2+2 to both sides, what we were really doing is making this into a perfect square trinomial which is something you might remember from our factoring formulas, and it's totally okay if you don't. You just need to know that this will always allow this to factor down into x+2, and that's going to be equal to some constant allowing us to just use the square root property. So let's go ahead and complete the square one more time here.

And here I have x2+8x+1=0. So let's start back at step 1, which is to simplify our equation to x2+bx=c. Now here I already have that leading coefficient of 1, but let's go ahead and move our constant over to the other side, which we can do by subtracting it. So it will cancel, leaving me with x2+8x=-1. Okay. So now we can move on to step 2 and add (b/2)2 to both sides. Now here, b is 8. So if I take 8 divided by 2 and then square it, that's going to give me 42 which is just 16. So I'm gonna go ahead and add 16 to both sides. Now when I add 16 to both sides step 2 is completed, and I can go ahead and move on to step 3 which is to factor this into x+b/22. Now we already said that our b/2 was this 8 / 2 or 4, so this will factor into x+42 and that's equal to -1 plus 16 which is just 15. Okay, so now that we've completed step 3, all we have left to do is to complete solving using the square root property, and I'm going to leave that up to you here. So that's all you need to know about completing the square, let's go ahead and get some more practice.

Solve the given quadratic equation by completing the square.

$x^2+3x-5=0$

• A. $x=-\frac32,x=\frac52$
• B. $x=-\frac32,x=\sqrt{29}$
• C. $x=\frac{-3+\sqrt{29}}{2},x=\frac{-3-\sqrt{29}}{2}$
• D. $x=\frac{3+\sqrt{29}}{2},x=\frac{3-\sqrt{29}}{2}$

Solve the given quadratic equation by completing the square.

$3x^2-6x-9=0$

• A. $x=3,x=-1$
• B. $x=3,x=1$
• C. $x=2,x=3$
• D. $x=-3,x=-4$

Quadratic Formula

Hey everyone. You just finished learning 3 different methods of solving quadratic equations, and you might be wondering how there's anything left and why we have yet another method to learn. But the quadratic formula that we're going to talk about now is really great because it's going to work for any quadratic equation. So even if you were to forget every single other method, as long as you remember the quadratic formula, you're going to be able to solve any quadratic equation that gets thrown at you. So let's go ahead and jump in. The quadratic formula is based on the standard form of a quadratic equation ax² plus bx plus c, and we're going to use a, b, and c in order to compute our solutions. So the quadratic formula is:

− b ± b 2 − 4 × a × c 2 × a .

Now that formula might look a little bit complicated right now and, unfortunately, it is something that you're going to have to memorize. But a way that I was able to memorize the quadratic formula was by using the quadratic formula song. Now, I'm not going to sing it for you right now, but it's gonna be something that you should search up on your own to commit this formula to memory. So when do we want to use the quadratic formula? Well, like I said, you can use the quadratic formula whenever you want, and some clues that you might want to use it are that you can't easily factor or you're just otherwise unsure what method to use.

Let's go ahead and take a look at an example here. So I have x² plus 2x minus 3 is equal to 0. Let's go ahead and take a look at our first step, which is to write our equation in standard form. Now it looks like my equation is already in standard form, all of my terms are on the same side in the descending order of power, so I can go ahead and move on to step 2. Now step 2 is simply to just plug everything into my quadratic formula, so let's go ahead and do that here. I'm first going to label a, b, and c in my equation so so that it's easy for me to just take them and plug them in. So in this case, I have an invisible one in front of that x² and that is my a, b is this positive 2, and then c is negative 3. Make sure that you're paying attention to the signs here. So plugging this in, the result is:

− 2 ± 2 2 − 4 × 1 × − 3 2 × 1 .

Now that we've plugged everything in, we've completed step 2, and all that we have left to do from here is algebra. We're just gonna compute and simplify our solutions now. So let's start with what's in our radical here and just simplify that. So this 2² is going to become a 4 minus 4 times 1 is just 4, so, really, that's just 4 times negative 3. I know that 4 times negative 3 is negative 12, and 4 minus negative 12, this becomes a plus stringByAppendingString,

Solve the given quadratic equation using the quadratic formula.

$3x^2+4x+1=0$

• A. $x=3,x=-1$
• B. $x=-\frac13,x=-1$
• C. $x=-3,x=-1$
• D. $x=\frac13,x=-1$

Solve the given quadratic equation using the quadratic formula. $2x^2-3x=-3$

• A. $x=\frac34+\frac{i\sqrt{15}}{4},x=\frac34-\frac{i\sqrt{15}}{4}$
• B. $x=\frac34+\frac{5i}{4},x=\frac34-\frac{5i}{4}$
• C. $x=\frac{3+\sqrt{15}}{4},x=\frac{3-\sqrt{15}}{4}$
• D. $x=3+i\sqrt{15},x=3-i\sqrt{15}$