Completing the Square - Video Tutorials & Practice Problems

Solving Quadratic Equations by Completing the Square

Hey, everyone, we just learned that whenever we have an equation in the form of X plus some number squared equals a constant, we can solve it using the square root property and simply square root 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 X squared plus six X is equal to negative seven. Well, I can actually take this equation and force it to be in that form X plus a number squared equals the constant 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 X squared plus six X plus something factor to X plus a number squared. And this is referred to as completing the square. So I'm gonna 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 one 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 X squared plus six X is equal to negative seven. Now let's check that we would even want to complete the square here. Well, my leading coefficient is definitely one and then B is six 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 one, which is to simplify our equation to this form squared plus BX equals C. Now, two notable things about this form are that my leading coefficient is one 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 X squared definitely already has a leading coefficient of one and my constant negative seven is already by itself on that right side. So step one is done and we can go ahead and move on to step two. Now, for step two, we're going to add A B over two squared to both sides. Now, this might seem really random. Why are we adding this B over two to both sides? But I'm gonna show you exactly how it works. Let's see what happens. So here B is six. So if I take 6/2 and then square it, that's gonna give me three squared, which is just nine. So let's go ahead and add that nine to both sides. So this gives me X squared plus six, X plus nine is equal to negative seven plus nine. So we've completed step two, we've added that B over two to both sides. Step three is going to be to factor this to X plus B over two squared. So this X squared plus six X plus nine actually factors perfectly into X plus B over two squared. Here, my B was six and B over two is just 3, 6/2. So this factors into X plus three squared and that's equal to negative seven plus nine, which is just two. So this is always how it's going to work. It is always going to factor perfectly into that B over two. And we've completed step three. 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 plus a number squared equals a constant, 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 one, which is to isolate our squared expression. Now, my squared expression is actually already by itself here that X plus three squared. So step one is done and I can go ahead and move on to step two, which is to take my positive and negative square root. So square rooting both sides, I am left with X plus three is equal to plus or minus the square root of two. So step two 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 three over which if I subtract three from both sides, I am then just left with X on my right or on my left side there. And then I have plus or minus a root two minus three. Now, we can make this look a little bit nicer by moving our negative three to the front there leaving me with negative three plus or minus the square root of two. And I'm done. Those are my solutions negative three plus or minus the square root of two. 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 over two squared 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 OK. If you don't, you just need to know that this will always allow this to factor down into X plus B over two squared. 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 X squared plus eight X plus one equals zero. So let's start back at step one, which is to simplify our equation to X squared plus BX equals C. Now, here I already have that leading coefficient of one, 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 X squared plus eight X is equal to negative one. OK. So now we can move on to step two and add B over two squared to both sides. Now, here B is eight. So if I take eight, divided by two and then square it, that's going to give me four squared, 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 two is completed and I can go ahead and move on to step three, which is to factor this into X plus B over two squared. Now, we already said that our B over two was this 8/2 or four. So uh this will factor into eight or X plus four squared and that's equal to negative one plus 16, which is just 15. OK. So now that we've completed step three, all we have left to do is to complete solving using the square root property. And I'm gonna 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. $x=-3,x=-4$