Completing the SquareBy L.D.

Slide 3: What is a perfect square? Slide 4: Making x2 + 12x + c a perfect square

trinomial Slide 8: Example Problems Slide 10: Completing the square to solve x2 +

6x = 7 Slide 18: Completing the square to solve x2 +

10x – 8 = 0

Table of Contents

It is a trinomial that is a perfect square in its factored

form. An example is x2 + 8x + 16, this trinomial is

(x + 4)2 in its factored form. The 2 exponent on the

end means that it is squared.

What is a perfect square?

Find the value that makes the expression a perfect square trinomial.

x2 + 12x + c

Problem 1

x2 + 12x + c

The first thing you do is make the form you will use for your problem by defining the things you already know. Since the very name of this presentation tells us the problem will be squared you can use a 2 exponent. You also can use and x at the beginning since you have x2 at the top. You also know that the problem will be positive due to the positive sign in front of the coefficient. The only thing left to find is what will fill the blank space.

(x + __ )2

Problem 1

x2 + 12x + c (x + __ )2

To fill the space you have to have that will add up to make 12 and make c be a number that it is easily to square. The simplest way to do this is to cut the coefficient in half to make sure that the number can be squared and still add to make the coefficient.

Problem 1

x2 + 12x + c (x + __ )2

This makes your answer (x + 6 )2.

To find out what c is, all you have to do is square the 6, c is 36. You can also get this answer by using F.O.I.L. on (x + 6 )2.

Problem 1

1. x2 + 14x + c

2. x2 – 8x + c

Example Problems

1. x2 + 14x + c (x + 7)2 c = 49 To do this you just use the basic form, (x + __)2 and slice

the coefficient in half to get your answer. You get c the same way we did in the last problem.

2. x2 – 8x + c (x - 4)2 c = 16

This problem is a bit different. Since there is a minus in front of the coefficient, the + in the form (x + __)2 will change to a minus. Besides that inconvenience, if we ignore the minus and just think of it as positive, we should easily get the answer we want and slide 4 in.

Example Problems

Complete the square to solve the equation

x2 + 6x = 7

Problem 2

Problem: Complete the square to solve the equation x2 + 6x = 71. Resist moving the seven to the other side.2. Add a space to help find the constant c1. x2 + 6x + ____ = 73. Get c by making the squared binomial form and then solve that form. The number gotten for c must be placed on the same side as the other constant.4. Solve the problem by using the previous square root method discussed on the blog and get the answers for x.

3. The problem will be solved step by step on the next few slides.

Condensed Instructions for Problem 2

x2 + 6x = 7

The first thing we have to do is resist moving the 7. At this stage of the problem, all of the things tainted with x stay on one side of the equation and everything untainted stays on the other. You may have to one day move the constant to the side away from the x side at this stage.

Problem 2

x2 + 6x = 7

After that is settled, you have to add a nice big space in the place the constant is supposed to be in a trinomial. There should be a plus sign next to it.

x2 + 6x + _____ = 7

Problem 2

x2 + 6x = 7 x2 + 6x + _____ = 7

Now that that is done, you ignore the 7 and think of the problem imaging the space as a c.

x2 + 6x + c You must now solve it like you solved problem

one to get (x + 3)2, making your c be 9.

Problem 2

x2 + 6x = 7 (x + 3)2

x2 + 6x + c (c = 9) = 7

Now you know that the 9 can’t be on the tainted side, so instead of placing our “c” on x side, you will place it on the side next to the seven and solve.

x2 + 6x = 9+ 7 x2 + 6x = 16

We now also know that your square problem (x + 3)2, is equal to 16, so our problem now looks like (x + 3)2 = 16

Problem 2

x2 + 6x = 7 (x2 + 6x)2 = 16

Now you use square roots to solve this like discussed on the previous post at my blog.

√(x + 3)2 = √16

x + 3 = ± 4

Problem 2

x2 + 6x = 7 x + 3 = ± 4

Now you use square roots to solve this like discussed on the previous post at my blog to get your final x answer.

√(x + 3)2 = √16

x + 3 = ± 4

x + 3 = 4 x + 3 = -4 -3 -3 -3 -3x = 1 x = -7

Problem 2

Complete the square to solve the equation

x2 + 10x – 8 = 0

Problem 3

x2 + 10x – 8 = 0

First you will scoot the 8 over to the untainted side and give the problem the right format.

x2 + 10x – 8 = 0 +8 +8 x2 + 10x = 8 After that you will find c (25) and make the squared

problem, which will be (x + 5)2 = 33

√(x + 5)2 = √33

x + 5 = ±√33

Problem 3

x2 + 10x – 8 = 0 x + 5 = ±√33

Unfortunately, the 33 can’t be squared. You will just have to leave it as is. You do have to move the 5 off the x side still though.

x + 5 = ±√33 -5 -5_____ x = -5 ±√33

You can go no further, so x = -5 ±√33 is the final answer.

Problem 3

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