Hey everyone. We just learned that whenever we have an equation in the form of x+number2 equals a constant, we can solve it using the square root property by simply taking 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 is equal to negative 7. Well, I can actually take this equation and force it to be in that form x+number2 = a 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 x2+6x+something factor to x+number2. 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 is equal to negative 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 over 2 squared to both sides. Now this might seem really random, why are we adding this b over 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 over 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 = negative 7 + 9. So we've completed step 2. We've added that b over 2 to both sides. Step 3 is going to be to factor this to x+b over 2 squared. So this x2+6x+9 actually factors perfectly into x+322. Here my b was 6 and b/2 is just 3, 6/2, so this factors into x+32 and that's equal to +2. So this is always how it's going to work. It is always going to factor perfectly into that b over 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+number2 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 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 is equal to plus or minus the square root of 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 right or on my left side there. And then I have plus or minus root 2 minus 3. Now we can make this look a little bit nicer by moving our negative 3 to the front there, leaving me with negative 3 plus or minus the square root of 2, and I'm done. Those are my solutions. Negative 3 plus or minus the square root of 2 and we've completely finished completing the square.
Now let's think about why that works for a second. When we added thisb over2 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 okay if you don't. You just need to know that this will always allow this to factor down into x+number2, 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 equals 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 is equal to negative one. Okay. So now we can move on to step 2 and add b over2 squared 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 over 2 squared. Now we already said that our b over 2 was this 8/2 or 4 so, this will factor into x + 4 squared and that's equal to negative 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.
