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Week Three Written Assignment Math 126

by KIKI
(Chicago,IL USA)











































Week Three Written Assignment



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 assignment must include (a) all math work required to answer the problems as well as (b) introduction and conclusion paragraphs.

Comments for Week Three Written Assignment Math 126

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Sep 29, 2011
Solve Quadratic Equations Using Indian Method
by: Staff

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Part III



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


Project 2:


Formula that yields prime numbers.

x² - x + 41

select at least five numbers; 0 (zero), two even, and two odd.

x values selected: 0, 6, 8, 11, 13

if x = 0

x² - x + 41

= 0² - 0 + 41

= 41


41 is a prime number

if x = 6

x² - x + 41

= 6² - 6 + 41

= 71


41 is a prime number

if x = 8

x² - x + 41

= 8² - 8 + 41

= 97

97 is a prime number


if x = 11

x² - x + 41

= 11² - 11 + 41

= 151

151 is a prime number


if x = 13

x² - x + 41

= 13² - 13 + 41

= 197

197 is a prime number



However the expression x² - x + 41 does yield composite numbers as well:

If x = 45

x² - x + 41

= 45² - 45 + 41

= 2021

2021 is a COMPOSIT number. Its factors are: 43 and 47

2021 can be factored into at least two smaller positive integers which not equal to 1.


Thanks for writing.

Staff
www.solving-math-problems.com



Sep 29, 2011
Solve Quadratic Equations Using Indian Method
by: Staff

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

Part II


Project 1:


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}


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

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Sep 29, 2011
Solve Quadratic Equations Using Indian Method
by: Staff


Part I

The question:

Week Three Written Assignment



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 assignment must include (a) all math work required to answer the problems as well as (b) introduction and conclusion paragraphs.


The answer:


Week Three Written Assignment


Following completion of your weekly readings, complete Projects #1 and 2 at the end of Chapter 7 in your textbook on page 331, Mathematics in Our World.

The paper should be one- to two-pages in length and 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 assignment must include (a) all math work required to answer the problems as well as (b) introduction and conclusion paragraphs.

Project 1:

An interesting method for solving quadric equations came from India. The steps are

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 term x2.
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 to the right 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.
try these:

A. x² - 2x - 13 = 0
B. 4x² - 4x + 3 = 0
C. x² + 12x - 64 = 0
D. 2x² - 3x - 5 = 0

Project 2:

Mathematicians have been searching for a formula the yields prime numbers.

One such formula was x² - x + 41.

Select some numbers for x, substitute then in the formula, and see if the prime numbers occur. Try to find a number for x that when substituted in the formula yields a composite number.

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Mar 19, 2012
Thank you for such a helpful post.
by: Anonymous

I want to thank you for I was doing my homework and having difficulty until I found your post. It was very helpful and I learned quite a bit from it. Once again, thank you so much for the help.

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