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Contents

- 1 Perfect Square Trinomials Missing Numbers
- 2 Perfect Square Trinomials Missing Numbers
- 3 Making a Perfect Square
- 4 Making a Perfect Square
- 5 Making Perfect Square Trinomials
- 6 Making Perfect Square Trinomials
- 7 Writing Squared Binomials
- 8 Writing Squared Binomials
- 9 Finding the Amount to Complete the Square
- 10 Finding the Amount to Complete the Square
- 11 Completing the Square with Equations
- 12 Completing the Square with Equations
- 13 Rewrite the Perfect Square Trinomials
- 14 Rewrite the Perfect Square Trinomials
- 15 Solving the Equation Using Completing the Square
- 16 Solving the Equation Using Completing the Square
- 17 Complete the Square Practice 1
- 18 Complete the Square Practice 1
- 19 Complete the Square Practice 2
- 20 Complete the Square Practice 2
- 21 Complete the Square Practice 3
- 22 Complete the Square Practice 3
- 23 Complete the Square Practice 4
- 24 Complete the Square Practice 4
- 25 Complete the Square Practice 5
- 26 Complete the Square Practice 5
- 27 Complete the Square Practice 6
- 28 Complete the Square Practice 6
- 29 Complete the Square Practice 7
- 30 Complete the Square Practice 7
- 31 Complete the Square Practice 8
- 32 Complete the Square Practice 8

Let's look at another way to solve quadratic equations, this time we're going to use a method called completing the square we'll see this method again when we look at conic sections so be sure you get familiar with it. Let's start by reviewing the perfect square trinomial patterns, what numbers would fill in each of these blanks be sure to indicate the appropriate sign in each box as well. So here you use a plus or a minus symbol, and here you'll put the number.

These were the missing signs and numbers for the perfect square trinomial patterns. Great work if you got all four of these correct. We know the first one needs to be positive 9, because this middle term is positive 18x. When we have the quantity a plus b squared, we know we'll get a squared plus 2ab plus b squared. So, this middle term is 2 times the multiplication of these two terms. Since we already have an x here, we know that we just need two times this number. So this number would have to be 9. We use similar reasoning to figure out the other three problems, and we just need to pay attention to what sign this is. This sign will indicate what sign we use in the parentheses. If it's a minus, we'll use minus. If it's a plus, we'll use plus.

We're going to use our knowledge of perfect square trinomials to help us complete the square. To complete the square means to change a polynomial into a perfect square trinomial. Let's consider this polynomial, x squared plus 4x. Now, let's use some tiles to help us visualize this polynomial. Here, we have a square that measures x by x. So its area is x squared, and here, we have 4 x's. We know that this side is also x and this side is just one unit long. So, 1 times x equals x, the area of this tile. So, if we wanted to have a perfect square, how many one tiles do you think we needed to add here to do that? In other words, what number should we add to our polynomial in order to get a perfect square?

We would need to add 4. If we added 4 then we would get a perfect square. We also know we can add 4 since 1 times 1 would equal 1 square here. We need 4 of these in total to get our perfect square. This trinomial is a perfect square since we can rewrite it as one. We can rewrite x squared plus 4x plus 4 as the quantity x plus 2 squared. In my head I'm picturing an actual square that has a side length of x plus 2, and another side length of x plus 2. This is why it's a perfect square. I can also see that my trinomial fits my perfect square pattern. We have a squared plus 2ab plus b squared. This means that we can rewrite a trinomial, as the quantity of a plus b squared.

What do you think we would need to add to each of these polynomials to make a perfect square? You want to think about drawing out a diagram like we did before, and then seeing how many one pieces you will need to complete the square. And don't be fooled by these negative signs here. Even if the x term is negative, we'll still add a positive amount to complete the square. So, what do you think for each of these? Take some time to think about it, and good luck.

We would need to add 9, 36, 4, and 25 to complete each of these squares. Great thinking if you found these numbers. For the first polynomial, we can think about having one block that's area is x squared, and then six other tiles that have an area of x. We take the 6x and we split three of the x's over here and three of the x's over here. So we'll have x squared plus six x's, and then we'll need to add at 9 one pieces to complete our perfect square. So, when thinking about completing the square, it's best if this b term is even. This allows us to split up the x's, so we have half of them here and half of them here. And then we can figure out how many one pieces we would need to add to complete our perfect square down here. For this first polynomial, it turns out that was 9. For the second one, we draw a similar figure as the one before, and we'll find out that we'll need to add 36 one tiles to complete the square. For this third polynomial, we said we had negative 4x here. So, we'll have our x squared like before, and then, we'll have negative 4 x's. So negative 2x and negative 2x, which makes negative 4x. So, if we were to lay out these tiles, we could see that we would need to add four more one pieces to complete the square. We know that this diagram also makes sense in terms of multiplication. We know x times x equals x squared, and x times negative 1 equals negative x. We have this negative x four times. Once here, once here, once here, and once here. And finally, for these 4 one pieces, we have a sidelength of negative 1 here and a sidelength of negative 1 here. So, negative 1 times negative 1 would equal a positive 1. We have this positive 1 four times, so we know we have to add 4 one tiles to complete our square.

So when we complete the square we add enough 1 pieces in order to make a perfect square. So, we can think about rewriting our polynomial x squared plus really be represented by a perfect square, with side lengths x plus 2, and x plus 2. We use our perfect square trinomial pattern to rewrite our trinomial as a perfect square. So, what do you think each of these trinomials would be written as a perfect square? Keep in mind when you enter your answer, you want to have something in parenthesis squared.

Here are the four perfect squares. Great job if you got those correct. We know we can use the perfect square pattern to figure out how to write each expression. We can look at the last term and take the square root of it to figure out what should go inside the parentheses. Also, since our middle term here is positive, we note that the sign that we'll use over here will also be positive. And these last two cases, the sign here for the middle term was negative, which means that the sign on the b term would be negative and our answer. If you're ever in doubt if you have a perfect square, simply multiply out this to see if you get this trinomial. It should work.

For these four examples, we were able to complete the square and turn each expression into a perfect square by adding a certain amount. The amount that we added in each case depended upon the b coefficient for the x term. In the first example, the b term was a 6 and we ended up having to add 9 to make a perfect square. In the second example, our b was 12 and we ended up adding 36 pieces to get a perfect square. In this third case, b was negative 4 and we added four pieces. And in the last case, b was negative 10 and we added 25. So, for any value of b, how much do you think we need to add in order to complete the square? In other words, see if you can look and find a pattern between these two numbers. How do we go from this b to the added amount? Now, you might think the answer would be b plus 3, but we know that can't be correct. Sure, if we add 3 here, we get 9. But if I added 3 to 12, I would get 15 and not 36. So, what expression could we write here that uses the letter b and gives us the appropriate value? The expression should work for each of the four cases. And be sure to take some time to think about it. This one's tough. If you get stuck, look for a hint in the instructor's comment.

This one is really tough, so don't worry if you struggled. The expression we want is the quantity b divided by 2 squared. For this first example, our b was equal to 6. So if we do 6 divided by 2 and square it, we would get 9. In the second case, b was equal to 12. So, 12 divided by 2 squared would be 6 squared, which is 36, the amount that we should add. In the third case, we had b was negative 4. So negative 4 divided by 2 equals negative 2, and then, we square it to get positive 4. Notice that in each case, we always added a certain amount. This is because even if b is negative, we end up squaring this value, so that way it turns out positive. To complete the square, we'll always add a positive amount, no matter if the b term is negative. And this also works for our last example. The b was negative 10, so we'll write that here, then we divided by 2, which gives us negative 5, and we square it to get negative 25. This is the powerful expression that we need in order to complete the square. If you found the expression, and that's pretty amazing. Again, this is a really hard pattern to find and generate on your own. Let's see how we can use this to solve some quadratic equations by completing the square.

Let's try using our knowledge of completing the square to solve this quadratic equation. Now, we want the left-hand side of the equation to contain a square so that we can take the square root of both sides. In other words, our goal is to make the left side look like a perfect square. And the right side to be some sort of number. Then, we could use our square root method that we learned before to continue solving. First, let's move this positive seven to the other side of the equation. We do this so we can start working towards getting a perfect square trinomial on the left. And a number on the right. Now that we're here, we want to add some amount to both sides of our equation. Remember, we can only perform the same operation to the left hand side of the equation and the right hand side of the equation, in order for our equation to remain balanced. So, what amount will we have to add here to complete the square? And since we add this amount here, remember, we'll add the same amount on the right.

To get a perfect square on the left. We'll need to add the quantity b over 2 squared to both sides of the equation. Here, our value of b is positive 6. We take 6, and we divide it by 2 to get 3. And then finally, we square that quantity to get 9. The amount we need to add to both sides of our equation.

So keep in mind our goal was to make the left hand side of the equation be a perfect square. So how do we write this left hand side, so it is a perfect square? What do you think goes here?

This trinomial is the quantity x plus 3 squared. We get the sign and the number of the binomial from the value b divided by 2. This 3 winds up inside of our parentheses or our perfect square. Now this is also the perfect time to check your work. If you square this or if we multiply this out, we should wind up with this trinomial. We can quickly write this out to get a product of x squared plus 6x plus 9. You could also probably do this in your head. This is a great way to check to see if you went from this step, to this step correctly.

Now that we've completed the square, we've actually reached our goal. We have a perfect square on the left, and a number on the right. This is an equation we know how to solve. We can simply take the square root of both sides, indicating the plus or minus here, and then solve for x. So you try finishing solving this equation. What would be the values for x? When you enter your answer, be sure to use the plus or minus symbol. Good luck.

Our solutions are negative 3 plus the square root of 2, or negative 3 minus the square root of 2. Nice work for finding those two answers. We start by taking the square root of both sides of our equation. So we'll have x plus 3 on the left equal to plus or minus root 2 on the right. Then to isolate x or to solve for x, we'll subtract 3 from both sides to get x is equal to negative 3 plus or minus root 2.

It's easiest to complete the square when solving a quadratic equation, if this quadratic is not factorable and if a is equal to 1 and b is even. We want b to be even because eventually we're going to add b over 2 squared to both sides of the equation. If b is even, then we know that this b over 2 squared ends up being an integer rather than a fraction. So, once we know a equals 1 and b is even. We can move this constant term to the other side of the equation. This will put the equation in this form. Next, we add b over 2 squared to both sides of our equation. This allows us to rewrite the left side as a perfect square. And then we can continue solving using our square root method. When solving the problems throughout this lesson. Use these steps as a guide to help you complete the square. Try solving this first practice problem. Move this negative 6 to the other side, and then complete the square and solve for x. Be sure you enter your answer using the plus or minus symbol. Good luck.

Here, the solutions were 2 plus the square root of 10 or 2 minus the square root of 10. Nice work if you got these two answers. So first, we add 6 to both sides of our equation, so we can set up the left side to complete the square. To complete the square, we're going to add b over 2 squared to both sides of our equation. The b here is negative 4. So, b over 2 would be negative 2 and b over 2 squared would be positive 4. Now, we're ready to write the left-hand side as a perfect square. Since our b term is negative, we know we're going to need x minus a number. It turns out it's x minus 2. Now you can either get that by thinking about what binomial squared would give us this trinomial or you can just pull it from this b over 2. This is negative 2, so we use that in our binomial. Now that we have our perfect square, we take the square root of both sides of the equation and we add two do both sides of our equation to solve for x. And we know we can't simplify the swuare root of ten since there's not a perfect square that divides into ten evenly

Try this second problem out. What do you think x would equal?

Here the solutions are 4 plus or minus the square root of 13. Great work if you found those answers. Again, we start by moving this constant term to the right side of our equation. So, we have x square, minus 8x equals negative 3. Next, we want to add the same amount to both sides of our equation, so that we can complete the square. Our b is negative 8. So, b over 2 squared equals positive our equation using x minus 4 squared. Now that we have a perfect square, we take the square root of both sides. To get x minus 4 is equal to plus or minus root 13. We add four to both sides to get x by itself, so 1x is equal to 4 plus or minus the square root of 13.

Here's a third problem on completing the square. What do you think that x could be in this equation? You'll want to use plus or minus in your answer, and be sure to simplify any radicals if you can.

First, we want to add 4 to both sides to get this polynomial by itself. Now, we're in a position to complete the square. We know our b is positive 4. So, b divided by 2 equals 2. And b over 2 squared is equal to 4. This is the amount we need to add to both sides of our equation in order to complete the square. So, we'll have positive 8 on the right and then we'll have x plus 2 for our perfect square on the left. Keep in mind, this positive 2, comes from our b over 2 term. This is positive 2, so this is positive 2. Another way to find this number, is the take the square root of this term and use the sign for the b term. The square root of 4 is 2 and we use the positive sign, since this sign is also positive. Next, we take the square root of both sides of our equation, to get x plus 2 equals positive or negative, 2 root 2. Finally we subtract 2 from both sides, to get our two answers for x.

Here's our fourth problem for completing the square. What values for x make this equation true? As a hint, this problem result in two imaginary solutions. Be careful when you simplify the negative square root. Good luck here.

First, we subtract 12 from both sides to get x squared plus 6x equals negative both sides of our equation. This allows us to rewrite the left side as a perfect square. X plus 3 squared. The right side just simplifies to negative 3. Next we take the square root of both sides of the equation to get x plus 3 equals plus or minus the square root of negative 3. This square root simplifies to plus or minus i root 3. It's an imaginary root here, since this radicand is negative. Finally, we subtract 3 from both sides to get x is equal to negative these two imaginary solutions.

Here's another problem for completing the square. What do you think are these solutions?

Here the solutions for x are 4 plus or minus 2i times root 5. Fantastic work if you found those solutions. This one wasn't so simple. First we subtract 36 from both sides, to get this equation. Next, we complete the square by adding b over negative 4. We square this number to get 16. Adding 16 to both sides of the equation, we get x minus 4 squared, this is our perfect square and negative 20 on the right. We take the square root of both sides of the equation, to get x minus 4 equals plus or minus the square root of negative 20. Then, we add 4 to both sides of the equation to get our solutions for x. We do need to do one more step. We need to simplify. The square root of negative 20 is going to be an imaginary number, since the radicand is negative. We can split up the square root of negative 20 into these three square roots. The square root of negative we leave it underneath the radical. These are our simplified solutions

How about this problem, what do you think the solutions would be for x? Write your answer here using the plus or minus symbol if there are two solutions.

Here are the solutions were 6 plus or minus the square root of 31. Great work if you found those two answers. We subtract 5 from both sides of the equation to get this. Next we identify the b so that way we can complete the square. The b value is negative 12, which means b over 2 is negative 6. And b over 2 squared is positive 36. We add this amount to both sides of our equations, so that way we can complete the square. We rewrite the left hand side as x minus 5 squared, and the right hand side as 31. Next, we take the square root of both sides of the equation to get x minus 6 equals plus or minus the square root of answers. Here, we can't simplify the square root of 31 since 31 is a prime number. The only factors of 31 are 31 and 1.

Here's our seventh practice problem. What do you think are the solutions for x? Try your best, and be careful when simplifying your radical. Good luck.

Here the answers were negative 3 plus or minus 5i, imaginary solutions. We start by simply completing the square, we have our x squared term here and our x term here. The b value is positive 6, so we find b over 2 which is 3 and then square this which is 9. We add 9 to both sides of our equation, so that way we can complete the square on this side. We rewrite our trinominal as the quantity x plus 3 squared, and then add these two together to get negative 25. We use our square root method to take the square root of both sides. So x plus 3 can equal positive or negative 5i. Keep in mind we have a negative radicant here, so the square root of 25 is 5 and the square root of negative 1 is i. And, finally, to solve for x, we simply subtract 3 from both sides to get our final answers.

Here's our last practice problem on completing the square. Notice that here we don't have an a value of 1. We have an a value of 3. So, what do you think you could do to solve here? Try taking an additional step here first and then try completing the square. When you think you have it, write your answer here.

Here the correct answers were 2 plus or minus i times root 6. That's amazing work if you got these answers correct. We want to start by dividing each term of our equation by 3. So dividing each of these by 3 we'll get x squared minus is one, and our b term is even. We subtract 10 from both sides to get x squared minus 4x equals negative 10. We complete the square by adding 4 to both sides of the equation. This gives us the quantity, x minus 2 squared equal to negative 6. Then, we take the square root of both sides of the equation to get x minus 2 equals positive or negative i root 6. We add 2 to both sides to solve for x. And these are the two solutions.