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Solving quadratic equations (equations with x2 can be done in different ways.  We will use two different methods.  What both methods have in common is that the equation has to be set to = 0.  For instance, if the equation was x2 – 22 = 9x, you would have to subtract 9x from both sides of the equal sign so the equation would be x2 – 9x – 22 = 0.

Solve by factoring: After the equation is set equal to 0, you factor the trinomial.

x2 – 9x – 22 = 0

(x-11) (x+2) = 0

Now you would set each factor equal to zero and solve.  Think about it, if the product of the two binomials equals zero, well then one of the factors has to be zero.

x2 – 9x – 22 = 0

(x-11) (x+2) = 0

x – 11 = 0        x + 2 = 0

+11    +11         -2       -2

x = 11      or       x = -2                   * Check in the ORIGINAL equation!

1) x2 - 5x - 14 = 0              2) x2 + 11x = -30                3) x2 - 45 = 4x

4) x2 = 15x - 56                 5) 3x2 + 9x = 54                  6) x3 = x2 + 12x

7) 25x2 = 5x3 + 30x             8) 108x = 12x2 + 216        9) 3x2 - 2x - 8 = 2x2

10) 10x2 - 5x + 11 = 9x2 + x + 83                11) 4x2 + 3x - 12 = 6x2 - 7x – 60

1) x2 – 3x – 108 = 0

2) x2 – 16x = - 55

3) x2 + 28 = 29x

4) x2 = x + 30

5) 6x2 = 54x – 120

6) x3 – 54x = 3x2

7) 7x2 + 14x = 560

8) 4x3 = -52x2 – 168x

9) 12x2 – 48x + 96 = 96

10) 15x3 – 28x = 105x2 – 28x

11) 4x2 – 96x = - 576

12) x2 – x = 5,256

Q3 Quiz 9 Review:

1) x2 – 14x = 15

2) x2 + 49 = -14x

3) 54 – 3x = x2

4) 8x2 + 192 = 88x

5) 2x3 + 30x2 + 112x = 0

6) 2x2 + 5x – 11 = 4x2 - 9x – 11

7) 3x2 – 11x + 31 = 7x2 – 11x – 33

8) 10x3 = 20x2 + 240x

9) x2 + 4x = 96

10) 7x6 + 35x5 = 350x4

11) 8x2 – 3x + 17 = 7x2 – 8x + 17

12) 5x2 – 4x – 13 = 4x2 + 3x + 17

1) x = {-1,15}

2) x = {-7}

3) x = {-9,6}

4) x = {3,8}

5) x = {-8,-7,0} 6) x = {0,7}

7) x = {-4,4}

8) x = {-4,0,6}

9) x = {-12,8}

10) x = {-10,0,5} 11) x = {-5,0}

12) x = {-3,10}

When ax2 + bx + c = 0

x =    -b  ±  √b2 – 4ac   .                                              2a

a is the coefficient of x2  b is the coefficient of x   c is the number (third term)

Notice the    ±    is what will give your two answers (just like you had when solving by factoring)

x2 – 9x – 22 = 0                       x =    -b  ±  √b2 – 4ac   . a = 1                                                             2a

b= - 9

c = -22                                     x =    -(-9)  ±  √ (-9)2 – 4(1)(-22)          -4(1)(-22) = 88                                                                        2(1)

x =   9  ±  √81 + 88                                                                    2

x=    9 ± √169   .                                                                          2

Split and do the   + side and - side

9 + 13                                      9 – 13                                        2                                                2

x = 11              or                     x = -2

* Check in the ORIGINAL equation!

31) 2x2 - 6x + 1 = 0             32) 3x2 + 2x = 3                33) 4x2 + 2 = -7x

34) 7x2 = 3x + 2                  35) 3x2 + 6 = 5x                 36) 9x - 3 = 4x2

1) The area of a rectangle is 80 cm2. The width is stated as x + 3 and the length is stated as x + 14. Find the value of x and then use x to find the perimeter.

8) The area of a rectangle is 117 in2. The width is 4 less than the length. Find the perimeter of the rectangle.

9) Find 5 negative consecutive integers such the product of the 2nd and the 5th is 504.

10) The area of a rectangle is 108 sq. feet.  The length is 3 more than the width.  Find the perimeter. (must use let)

11) The area of a rectangle is 176 sq. feet.  The width is 5 less than the length.  Find the perimeter. (must use let)

12) The area of a rectangle is 135 sq. feet.  The length is 6 more than the width.  Find the perimeter. (must use let)

13) Find 5 positive consecutive integers such the product of the 1st and the 5th is 221.

14) Find 4 consecutive negative odd integers such that the product of the 2nd and 4th is 357.

Solving Perfect Squares

When the square root of a constant, variable or polynomial results in a constant the expression is a perfect square. Some examples are 49, 64, and .

Example:

Steps:

1.

1. Be sure that the binomial is fully isolated.

2. Take the square root of both sides.

3. Split the solution into two equations.

4. Solve both equations for x.

More Examples:

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Completing the Square

Completing the square is another method that is used to solve quadratic equations. This method is especially helpful when the quadratic equation cannot be solved by simply factoring.

***Remember the standard form for a quadratic equation is: ax2 + bx + c = 0.***

Example:

Steps:

1.

1. Be sure that the coefficient of the highest exponent

is 1. If it is not divide each term by that value to

create a leading coefficient of 1.

2. Move the constant term to the right hand side.

3. Prepare to add the needed value to create a perfect

square trinomial. Be sure to balance the equation.

4. To create the perfect square trinomial:

a) Take

b) Add that value to both sides of the equation.

5. Factor the perfect square trinomial.

6. Rewrite the factors as a squared binomial.

7. Take the square root of both sides.

8. Split the solution into two equations

9. Solve for x.

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Solving Perfect Squares

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Completing the Square

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Proportions and Percents

Proportions:

A proportion is a statement that two ratios are equal.  When trying to solve proportions we use the Cross Products Property of Proportions.

A      =     C                          A(D) = B(C)

B              D

Example:

6__   =     x__                                                 x + 5__   =       1.5___

11           121                                                    12                    6