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mat 126

by lanitra
(Hyattsville)











































For Project #1, complete all 6 steps (a-f) as shown in the example. For Project #2, please select at least 5 numbers; 0 (zero), 2 even and 2 odd. Make sure you organize your paper into separate project



(a) Move the constant term to the right side of the equation.
(b) Multiply each term in the equation by four times the coefficient of the X2 term
(c) Square the coefficient of the original x term and add it to both sides of the equation.
(d) Take the square root of both sides.
(e) Set the left side of the equation equal to the positive square root of the number on the right side and solve for x.
(f) Set the left side of equation equal to the negative square root of the number on the right side of the equation and solve for x.


Example : solve x2 + 3x - 10 = 0


X2 + 3x = 10

4x2 + 12 x = 40

4x2 + 12x + 9 = 40 + 9

4x2 +12x + 9 = 49

2x + 3 = _+ 7 ( the plus sign should sit on top of the - sign)



2x + 3 = 7 2x +3 = -7

2x = 4 2x =-10

X=2 x=-5

Solve these show work.

(a) X2 _ 2x - 13= 0
(b) 4 x2 - 4x + 3 =0
(c) X2 + 12x - 64= 0
(d) 2x2 -_ 3x - 5 =0


Comments for mat 126

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Jun 28, 2011
Solve Equations
by: Staff

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Part III


(d) 2x² - 3x - 5 = 0

Part (d) will also be solved by “COMPLETING THE SQUARE”.


Add 5 to each side of the equation to move the 5 to the right side of the equation

2x² - 3x - 5 + 5 = 0 + 5

2x² - 3x = 5


Divide each side of the equation by 2 to remove the 2 from as the coefficient of 2x² term on the left side of the equation


(2x² - 3x)/2 = 5/2

x² - 3x/2 = 5/2



Divide the coefficient of - 3x/2 by 2

(- 3/2)/2 = - ¾

Square the result

(- ¾)² = 9/16

Add this value to each side of the equation. This means that you will add 9/16 to each side of the equation.

x² - 3x/2 + 9/16= 5/2 + 9/16

x² - 3x/2 + 9/16= 40/16 + 9/16

x² - 3x/2 + 9/16= 49/16

As before, THE REASON for moving the “5” to the right side of the equation, then dividing each side of the equation by 2, and then adding “9/16” to each side of the equation is: CHANGE THE LEFT SIDE OF THE EQUATION SO IT CAN BE FACTORED AS A PERFECT SQUARE:

x² - 3x/2 + 9/16= 49/16

(x - ¾)(x - ¾) = 49/16


(x - ¾)² = 49/16


(x - ¾)² = (7/4)²


Take the square root of each side of the equation


√(x - ¾)² = ±√(7/4)²

x - ¾ = ±(7/4)

x - ¾ + ¾ = ¾ ±(7/4)

x = ¾ ±(7/4)


1st value of x = ¾ plus 7/4

x = ¾ + (7/4)

x = 10/4

x = 5/2


2nd value of x = ¾ minus 7/4

and x = ¾ - (7/4)

and x = - (4/4)

and x = - 1

x ∈{5/2, - 1}



the final answer to part (d) is: x = 5/2 and x = - 1, or x ∈{5/2, - 1}





Thanks for writing.

Staff
www.solving-math-problems.com


Jun 28, 2011
Solve Equations
by: Staff

-----------------------------------------------------

Part II


(b) 4x² - 4x + 3 = 0

This equation is similar to part (a). We will solve it by “COMPLETING THE SQUARE”.


Subtract 3 from each side of the equation to move the 3 to the right side of the equation


4x² - 4x + 3 - 3 = 0 – 3

4x² - 4x = - 3

4(x² - x) = - 3

Divide each side of the equation by 4 to remove the 4 from the left side of the equation

4(x² - x) = - 3

4(x² - x)/4 = - ¾

x² - x = - ¾


Divide the coefficient of -x by 2

-1/2 = -1/2

Square the result

(-1/2)² = ¼

Add this value to each side of the equation. This means that you will add ¼ to each side of the equation.

x² - x = - ¾

x² - x + ¼ = - ¾ + ¼

x² - x + ¼ = - ½

THE REASON for moving the “3” to the right side of the equation, dividing each side of the equation by “4”, and then adding “¼” to each side of the equation is: CHANGE THE LEFT SIDE OF THE EQUATION SO IT CAN BE FACTORED AS A PERFECT SQUARE:

x² - x + ¼ = - ½

(x - ½)(x - ½) = - ½

(x - ½)² = - ½


Take the square root of each side of the equation

√(x - ½)² = ±√(-½)

x - ½ = ±√(-½)


Add ½ to each side of the equation to remove the ½ from the left side of the equation. This leaves the variable x as the only term on the left side of the equation.

x - ½ + ½ = ½ ±√(-½)

x = ½ ±√(-½)

x = ½ ± i√(½)

1st value of x = ½ plus the square root of (-1/2)


x = ½ + 0.70710678118654752440084436210485 i

x = 0.5 + 0.707107 i


2nd value of x = ½ minus the square root of (-1/2)

x = ½ - 0.70710678118654752440084436210485 i

x = 0.5 - 0.707107 i



the final answer to part (b) is: x = ½ ± i√(½), or x ∈{0.5 + 0.707107 i, 0.5 - 0.707107 i }
(note that the value of x includes an imaginary number)



(c) x² + 12x - 64 = 0

Part (c) will also be solved by “COMPLETING THE SQUARE”.


Add 64 to each side of the equation to move the 64 to the right side of the equation

x² + 12x - 64 + 64 = 0 + 64

x² + 12x = 64



Divide the coefficient of 12x by 2

12/2 = 6

Square the result

6² = 36

Add this value to each side of the equation. This means that you will add 36 to each side of the equation.

x² + 12x + 36 = 64 + 36

x² + 12x + 36 = 100


THE REASON for moving the “64” to the right side of the equation, and then adding “36” to each side of the equation is: CHANGE THE LEFT SIDE OF THE EQUATION SO IT CAN BE FACTORED AS A PERFECT SQUARE:

x² + 12x + 36 = 100

(x + 6)(x + 6) = 100

(x + 6)² = 100


Take the square root of each side of the equation

√(x + 6)² = ±√10²

x + 6 = ±10


Subtract 6 from each side of the equation to remove the 6 from the left side of the equation. This leaves the variable x as the only term on the left side of the equation.

x + 6 - 6 = -6 ±10


1st value of x = -6 plus 10

x = -6 + 10

x = 4


2nd value of x = -6 minus 10

x = -6 - 10

x = -16

x ∈{4, -16}

the final answer to part (c) is: x = 4 and x = -16, or x ∈{4, -16}

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Jun 28, 2011
Solve Equations
by: Staff


Part I

The question:

by LaNitra
(Hyattsville)

For Project #1, complete all 6 steps (a-f) as shown in the example. For Project #2, please select at least 5 numbers; 0 (zero), 2 even and 2 odd. Make sure you organize your paper into separate project



(a) Move the constant term to the right side of the equation.
(b) Multiply each term in the equation by four times the coefficient of the X2 term
(c) Square the coefficient of the original x term and add it to both sides of the equation.
(d) Take the square root of both sides.
(e) Set the left side of the equation equal to the positive square root of the number on the right side and solve for x.
(f) Set the left side of equation equal to the negative square root of the number on the right side of the equation and solve for x.


Example : solve x2 + 3x - 10 = 0


X2 + 3x = 10

4x2 + 12 x = 40

4x2 + 12x + 9 = 40 + 9

4x2 +12x + 9 = 49

2x + 3 = _+ 7 ( the plus sign should sit on top of the - sign)



2x + 3 = 7 2x +3 = -7

2x = 4 2x =-10

X=2 x=-5

Solve these show work.

(a) X2 _ 2x - 13= 0
(b) 4 x2 - 4x + 3 =0
(c) X2 + 12x - 64= 0
(d) 2x2 -_ 3x - 5 =0


The answer:


Solve these show work.


(a) x² - 2x - 13 = 0

we will solve this equation by “COMPLETING THE SQUARE”


add 13 to each side of the equation to move the 13 to the right side of the equation

x² - 2x - 13 + 13 = 0 + 13

x² - 2x = 13



Divide the coefficient of -2x by 2

-2/2 = -1

Square the result

(-1)² = 1

Add this value to each side of the equation. This means that you will add 1 to each side of the equation.

x² - 2x + 1= 13 + 1

x² - 2x + 1= 14


THE REASON for moving the “13” to the right side of the equation, and then adding “1” to each side of the equation is: CHANGE THE LEFT SIDE OF THE EQUATION SO IT CAN BE FACTORED AS A PERFECT SQUARE:

x² - 2x + 1= 14

(x - 1)(x - 1) = 14

(x - 1)² = 14


Take the square root of each side of the equation

√ (x - 1)² = ±√(14)

x - 1 = ±√(14)


Add 1 to each side of the equation to remove the 1 from the left side of the equation. This leaves the variable x as the only term on the left side of the equation.

x - 1 + 1 = 1 ± √(14)

x = 1 ± √(14)


1st value of x = 1 plus the square root of 14

x = 1 + √(14)
x = 1 + 3.7416573867739413855837487323165

x = 4.7416573867739413855837487323165

x = 4.741657


2nd value of x = 1 minus the square root of 14

x = 1 - √(14)

x = 1 - 3.7416573867739413855837487323165

x = -2.7416573867739413855837487323165

x = -2.741657


x ∈{4.741657, -2.741657}


the final answer to part (a) is: x = 1 ± √(14), or x ∈{4.741657, -2.741657}

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