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Math Project

by Nique
(Rockford IL)











































Project #1

Solve the following quadratic equations

    (a) : x² - 2x - 13 = 0

    (c) : x² + 12x - 64 = 0

Use the “Indian Method” to solve for “x”.


Project #2

please select at least five numbers; 0 (zero), two even, and two odd. Make sure you organize your paper into separate projects.


Refer to the “Projects” section on page 331 of Mathematics in Our World.

Comments for Math Project

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Aug 09, 2011
Math Project
by: Staff

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Part II


check the solution by substituting the two numerical values of x into the original equation

for x = -2.74166

x² - 2x - 13 = 0

(-2.74166)² - 2(-2.74166) - 13 = 0

7.51669 + 5.48332 - 13 = 0

13 - 13 = 0, OK


for x = 4.74166

x² - 2x - 13 = 0

(4.74166)² - 2(4.74166) - 13 = 0

22.4833 - 9.48332 - 13 = 0

13 - 13 = 0, OK



Equation (c) : x² + 12x - 64 = 0

(a) Move the constant term to the right side of the equation.

x² + 12x - 64 = 0

x² + 12x - 64 + 64 = 0 + 64

x² + 12x + 0 = 0 + 64

x² + 12x = 0 + 64

x² + 12x = 64


(b) Multiply each term in the equation by four times the coefficient of the x squared term.

The coefficient of the x² term is 1.

x² + 12x = 64

(4 * 1) * (x² + 12x = 64)

(4) * (x² + 12x = 64)

(4)*x² + (4)*(12x) = (4)*(64)

4x² + 48x = 256



(c) Square the coefficient of the original x term and add it to both sides of the equation.

The coefficient of the original x term is 12.

(12)² = 144

4x² + 48x = 256

4x² + 48x + 144 = 256 + 144

4x² + 48x + 144 = 400


(d) Take the square root of both sides.

4x² + 48x + 144 = 400

Sqrt(4x² + 48x + 144) = Sqrt(400)

Sqrt(2x + 12)² = Sqrt(20²)

Sqrt(2x + 12)² = ±20

2x + 12 = ±20

(to simplify both sides of the equation, divide each side of the equation by 2)

(2x + 12)/2 = ±20/2

x + 6 = ±10


(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.

x + 6 = 10

x + 6 - 6 = 10 - 6

x + 0 = 4

x = 4


(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.

x + 6 = -10

x + 6 - 6 = -10 - 6

x + 0 = -16

x = -16


the final solution is: x ∈ {-16, 4}


check the solution by substituting the two numerical values of x into the original equation

for x = -16

x² + 12x - 64 = 0

(-16)² + 12(-16) - 64 = 0

256 - 192 - 64 = 0

256 - 256 = 0, OK


for x = 4

x² + 12x - 64 = 0

(4)² + 12(4) - 64 = 0

16 + 48 - 64 = 0

64 - 64 = 0

0 = 0, OK



Thanks for writing.

Staff
www.solving-math-problems.com



Aug 09, 2011
Math Project
by: Staff


Part I

The question:

by Nique
(Rockford IL)

Following completion of your weekly readings, complete the exercises in the “Projects” section on page 331 of Mathematics in Our World.

You should be concise in your reasoning. For Project #1, work only equations (a) and (c), but complete all 6 steps (a-f) as shown in the example.

For Project #2, please select at least five numbers; 0 (zero), two even, and two odd. Make sure you organize your paper into separate projects.


The answer:

Equation (a) : x² - 2x - 13 = 0
Equation (c) : x² + 12x - 64 = 0


The Indian Method for solving a quadratic equation is:

(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 x squared 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.
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Equation (a) : x² - 2x - 13 = 0

(a) Move the constant term to the right side of the equation.

x² - 2x - 13 = 0

x² - 2x - 13 + 13 = 0 + 13

x² - 2x + 0 = 0 + 13

x² - 2x = 0 + 13

x² - 2x = 13


(b) Multiply each term in the equation by four times the coefficient of the x squared term.

The coefficient of the x² term is 1.

x² - 2x = 13

(4 * 1) * (x² - 2x = 13)

(4) * (x² - 2x = 13)

(4)*x² + (4)*(- 2x) = (4)*(13)

4x² - 8x = 52
(c) Square the coefficient of the original x term and add it to both sides of the equation.

The coefficient of the original x term is -2.

(-2)² = 4

4x² - 8x = 52

4x² - 8x + 4 = 52 + 4

4x² - 8x + 4 = 56


(d) Take the square root of both sides.

4x² - 8x + 4 = 56

Sqrt(4x² - 8x + 4) = Sqrt(56)

Sqrt(2x - 2)² = Sqrt(2²*2*7)

Sqrt(2x - 2)² = Sqrt(2²)*Sqrt(14)

(2x - 2) = 2*Sqrt(14)

(to eliminate the 2’s on both sides of the equation, divide each side of the equation by 2)

(2x - 2)/2 = (2/2)*Sqrt(14)

(x - 1) = (1)*Sqrt(14)

x - 1 = (1)*Sqrt(14)


(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.

x - 1 = +Sqrt(14)

x - 1 + 1 = Sqrt(14) + 1

x + 0 = Sqrt(14) + 1

x = 1 + Sqrt(14)

x = 1 + 3.74166
x = 4.74166


(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.

x - 1 = -Sqrt(14)

x - 1 + 1 = -Sqrt(14) + 1

x + 0 = -Sqrt(14) + 1

x = 1 - Sqrt(14)

x = 1 - 3.74166

x = -2.74166


the final solution is: x ∈ {-2.74166, 4.74166}
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