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math week 3

by Lisa
(MD)











































ASSIGNMENTS


PROJECT #1. Work only equations (a) and (c), but complete all 6 steps (a-f) as shown in the example. An interesting method for solving quadratic equations came from India. The steps are;
You should be concise in your reasoning.

(a). Move the constant term to the right side of the equation.
(b). Multiple each term in the equation by four times the coefficient of the x^2 term.
(c). Square the coefficient of the original x term and add it t 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 the equation equal to the negative square root of the number on the right side of the equation and solve for x..

Equations to solve;

(a). x^2 – 2x – 13 = 0

(c). x^2 + 12x – 64 = 0


PROJECT #2. Please select at least five numbers; 0 (zero), two even, and two odd of your choice. Make sure you organize your paper into separate projects. Use the same steps to solve project #2 as well.

Comments for math week 3

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Jun 29, 2011
Quadratic Equations: Completing the Square
by: Staff

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Part II

(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}




Project #2

There is not enough information to complete this project.



Thanks for writing.

Staff
www.solving-math-problems.com



Jun 29, 2011
Quadratic Equations: Completing the Square
by: Staff


Part I

The question:

by Lisa
(MD)

ASSIGNMENTS


PROJECT #1. Work only equations (a) and (c), but complete all 6 steps (a-f) as shown in the example. An interesting method for solving quadratic equations came from India. The steps are; You should be concise in your reasoning.

(a). Move the constant term to the right side of the equation.
(b). Multiple each term in the equation by four times the coefficient of the x^2 term.
(c). Square the coefficient of the original x term and add it t 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 the equation equal to the negative square root of the number on the right side of the equation and solve for x..

Equations to solve;

(a). x^2 – 2x – 13 = 0

(c). x^2 + 12x – 64 = 0

PROJECT #2. Please select at least five numbers; 0 (zero), two even, and two odd of your choice. Make sure you organize your paper into separate projects. Use the same steps to solve project #2 as well.


The answer:

Project #1

(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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