(Click one to begin)

Completing the Square

Quadratic equation

Factoring (foil)

2x2 + 10x + 8=-4

SET THE EQAUTION EQUAL TO ZERO

2x2+10x+8=-4

To set the equation equal to zero simply add four to both sides of the equation.

2x2 + 10x + 12= 0

Click for answer

TAKE OUT THE GREATEST COMMON FACTOR (GCF)

2x2 + 10x + 12=0To complete the next step of this method you take out the Greatest Common Factor (GCF) or largest number that is a factor of every term in the equation. For this equation the GCF is 2, because it is the largest factor of ALL the terms.

2(x2 + 5x + 6)= 0Click for answer

SET UP TWO SETS OF PARENTHESIS ( ) ( )…

( ) ( )

… Easy enough

Step 3 continued

The next thing that you need to do to solve the equation is learn how to “FOIL”. To set up your “FOIL” equation you place the letters of the word FOIL to the corresponding places shown below.

( ) ( )F L F L

O I I O

CONTINUE FOILING

The next step in “FOILing” the equation is to take the first term in the trinomial and take half of the exponent and put the variable and exponent in the “F” places in the equation. The trinomial in this equation is (x2 + 5x + 6), so the first term would be x2. Half of the exponent (2) would be 1, so you would put x1 (x) in the “F” places.

( x ) (x ) F L F L

O I O I

Click for answer

WHAT OPERATION TO USE?

After you have the first term in each of your Parenthesis you need to put an operation, either addition or subtraction, in the equation. You need to know whether or not the operations will be the same, and what they will be, addition or subtraction. To do this you look at the second sign, or operation, in the trinomial. If it is positive (addition) the signs will be the same (whatever the first sign is), if it is negative (subtraction) the signs are different. In our example problem the trinomial is (x2 + 5x + 6), so the second sign is addition. We now know that the signs will be the same. The first sign is positive, so we know that both sets of parenthesis will have addition for a sign.

(x+ ) (x+ )Click for answer

FIND THE SECOND TERM

Now that you have the first term, and operation of both sets of parenthesis, you need to find the second term for each set. To do this you have to list the factors of the last term in the trinomial. In our equation the last term in the trinomial is 6. (x2 + 5x + 6).

61 6

2 3

Click for answer

WHAT TO DO WITH THE FACTORS?

If the second sign in your trinomial is addition, you find the pair of factors that add to get the second term in the trinomial. If the second sign is subtraction, you find the pair of factors that subtract to get the second term. The second term in our problem is addition, so you would have to find the factor pair of six, that adds to get five. If there are no factor pairs that add, or subtract to equal B, the equation is PRIME.

(x2 + 5x + 6)6

1 + 6 = 7

2 + 3 = 5

3 and 2 are the factor pair you want, because they add to equal 5.

Click for answer

(Continued)

Leading coefficient other than one

If the leading coefficient of an equation is a value other than one, you have to find the factors of (a) (first coefficient), and (c) (third coefficient). You multiply the factors of (a) by the factors of (c), and find a combination that either add or subtract to equal (b) (depending on the sign). To determine where the factors a placed in the two sets of parenthesis you find the greater product of the two sets. You take the last value of the pair of factors with the greatest product, and place it in the last spot in the parenthesis that

has a corresponding sign with that in the first spot in the equation.

0262 y4y

4

1 4

2 2

2

1 2

Factors of 4 Factors of 2

(1) (2) = 2

(4) (1) = 4

2 and 4 add to equal 6

(4y+2)(y+1)

2

1-1,-y

Click for answer

FINISHING UP

This factor pair is going to be the last thing you put into your set of parenthesis to solve the problem. Once you have found your factor pair you have to decide which value goes into which parenthesis. To do this you have to look at the signs that you put in your parenthesis in step 5. If the signs are the same, place the factor pairs in the “L” spots. If the signs are different, place the greatest factor of the pair in the parenthesis that contains the same sign as the first of the trinomial. If the first sign of the trinomial is (–) and your factor pair is (6,3) then you place the 6 in the parenthesis with the (–) sign. In our problem the signs are both addition, so you place one value in each parenthesis.

2(x + 3)(x + 2) Click for answer

Finding x’s values

Now that you have your new equation 2(x+3)(x+2) you need to find out what x equals. To do this you set the equation equal to zero, and x will then equal the opposite of whatever is being added to it, or subtracted from it, in each parenthesis.

2 (x + 3) (x + 2) = 0

x= -3, -2 This is your final answer

Click for answer

To check you answer you go use the “FOIL” set up, and multiply the value under each letter in the first parenthesis, by the corresponding value of the same letter in the second parenthesis.

(x + 3) (x + 2) F L F L

O I I OF: (x)(x)= x2

O: (x)(2)= 2x

I: (3)(x)= 3x

L: (3)(2)= 6

Once you have these values, you add them all together. If the equation you get corresponds to the original equation your answer is correct.

Click for answer

x2 – 7x – 10= 0

ax2 + bx + c = 0

MOVE THE CONSTANT

x2 – 7x – 10 = 0The first step in solving an equation using this method is to move the constant to the right side of the equal sign. The terms with variables stay on the left side of the equal sign. In this equation, you move the constant to the right side of the equal sign by adding it (10) to both sides.

x2 – 7x = 10

Click for answer

LEADING COEFFICIENT

In some equations, the leading coefficient will be a value other than one. In the equation 2x2 +4x=8, the leading coefficient is 2. If you have an equation like this, you divide every term by the leading coefficient.

2x2 + 4x = 8(Divide every term by 2)

x2 + 2x = 4Click for answer

USING “B”

x2 – 7x = 10

The next step in solving this equation involves using the formula for completing the square, which is, ax2 + bx + c= 0. The next thing that you would do is to take (1/2 b)2 and add it to both sides of the equation. In our equation b is –7. Half of 7 is (7/2). (7/2)2 is (49/4)

Click for answer

4

4910

4

49x7x2

SIMPLIFY

The next step you take is to simplify the right side of the equation. To do this in this equation, you would add 10 to the quantity of (49/4). In order to add, the two values must have the same denominator. You could change 10 to (40/4). This would give you common denominators so you could then simply add the terms.

Click for answer

4

4910

4

49x7x2

4

89

4

49x7x2

4

49

4

40

4

49x7x2

REWRITE

Once you have simplified the right side of the equation, the next thing to do is rewrite the left side of the equation as a binomial squared. To do this, you take the quantity of x plus or minus (depending on what the first sign is) half of b and square that quantity. The first sign in this equation is subtraction, so we would subtract half of b from x and square the quantity. Click for answer

4

89

4

49x7x2

2

2

7x

SQUARE ROOT

After you rewrite the left side of the equation, and the equation looks similar to the one shown above, the next thing to do is take the square root of both sides of the equation. The left side of the equation is a quantity squared, so the square root of it is simply that quantity. For the right side of the equation, you take the square root of the numerator and the denominator and keep it as a fraction. Since the square root of a term is a positive and negative value, the left side will equal + or - the right.

4

89

2

7x

2

2

89

2

7x

Click for answer

SOLVE

The final step in solving the equation using completing the square is to solve for x. To do this, the first thing you need to do, is get x alone on one side of the equation. Then you would move the other values on the same side as x to the other side of the equation. In this problem you would ADD (7/2) to both sides of the equation. You would then have x alone and could simplify to get your final answer. In some equations the number inside the square root sign will be negative. If your equation has an answer like that, you could leave it as it is, or write “no rational roots”. You could write this because the square root of a negative number is not rational.

(SIMPLIFY)

2

89

2

7x

2

89

2

7x

2

897x

Click for answer

a)(2

)c)(a(4bb 2

02x)3(x4 2

Learning the formula

a2

)c)(a(4bb 2

Using the Quadratic equation (shown above) you can set up your equation to work in the quadratic formula (shown below).The example equation is set up to work in the quadratic formula.

0cbxax2 02x)3(x4 2

x=

Click for answer

Setting up the formula

02)3(4 2 xxNow that the equation is set up in the form of the correct form you can substitute the values of your equation for the corresponding variables in the quadratic formula (shown below)

02)3(4 2 xx a=4 c=2

b=(-3)

a

cabb

2

))((42 = )4(2

)2)(4(4)3(3 2 x= x=

Click for answer

Simplify

)4(2

)2)(4(4)3(3 2

After you have put all the values in the equation into the quadratic formula the next step is to simplify (using order of operations).

x=

)4(2

)2)(4(493 = 8

3293 =

8

233 x=

Click for answer

Finishing up

In this equation the final answer contains the square root of a negative number. The answer to this would be not be rational, so therefore your final answer could be “no rational roots” or leave it as it is. This is only the case in problems where you have a square root of a negative number. If your answer contains the square root of a positive number, you could simplify (if it contains a perfect square) or leave it as is.

8

233 x=

3

84 x= Can be reduced to:

3

224 (Click for answer)

Menu

Practice problems

Click for problem 1

Solve by factoring

6672 yyANSWERS

(choose one)

4,3y

4,3y

4,3 y

4,3 y

6672 yy

01272 yy

0)4)(3( yy

4,3y

Solve by using the quadratic formula

0352 2 xx

1,2

3-x

1,2

3x

2

1,3-x

1,2

3x

0352 2 xx

)2(2

)3)(2(455 2 x

4

24255 x

4

15x

4

1

4

5

x

1,2

3

x

Solve by completing the square

063242 zz

3,21z2

3,21z

3,21z 3,3

21z

063242 zz

63242 zz81144242 zz

81122

z912 z

912z

3,21z

Solve by factoring

08102 2 ww

-1,-4w

3,4

1-w

1,4

1w

Prime

08102 2 ww

0452 2 ww

0412 ww

4,1w

Solve by using the quadratic formula

132 xx

Answer

2

133x

132 xx

)1(2

)1)(1(493

2

)4(93

2

133

Solve by completing the square

012 xx

Answer

2

51x

4

5

4

12 xx12 xx02 1-xx

4

52

2

1-x

2

5

2

1x2

5

2

1 x

2

51x

Solve by factoring

027x3x2

2,3

1x

2

1,3 x 2,

3

1x

Prime

027x3x2

01)2)(3x(x

2,3

1x

089gg2 4

Solve by using quadratic formula

Answer

8

47-9-g

(no rational roots)

089gg2 4

)4(2

4(4)(8)-819-

8

128-819-

8

47-9-

Solve by completing the

square

0433 2 rr

Answer

6

393 r (no rational roots)

0433 2 rr

433 2 rr

3

42 rr

12

13

4

1rr2

12

132

2

1-r

32

13

2

1r

6

393 r

0132 xx

Prime

1,3x 1,3x

1,3

1x

The answer to this equation is PRIME (b), because no factors of 1, add to equal 3.