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Solving math equations - Mathematics in Our World - Projects - Page 397
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Solving math equations - Mathematics in Our World - Projects - Page 397







































Following completion of your weekly readings, complete the exercises in the “Projects” section on page 397 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.

Your introduction should include three to five sentences of general information about the topic at hand.
The body must contain a restatement of the problems and all math work, including the steps and formulas used to solve the problems.
Your conclusion must comprise a summary of the problems and the reason you selected a particular method to solve them. It would also be appropriate to include a statement as to what you learned and how you will apply the knowledge gained in this exercise to real-world situations.

The Steps in Solving Math Equations
This weeks exercises included two different projects listed on page 397 of Mathematics in Our World (Bluman, 2011):
1. For project #1 we are to solve equations ( a ) and ( c ) using all 6 steps listed in the example.
2. For project #2 we are to select at least five numbers consisting of 0, two even, and two odd.
Project #1
The basis of this project actually comes from an interesting method for solving quadratic equations. This method originated from India. Below is the list of the step by step instructions I used from the method derived in India to solve equations ( a ) : x² - 2x - 13 = 0 and ( c ) : x² + 12x - 64 = 0
1. Move the constant term to the right side of the equation.
2. Multiply each term in the equation by four times the coefficient of the x squared term.
3. Square the coefficient of the original x term and add it to both sides of the equation.
4. Take the square root of both sides.
5. Set the left side of the equation equal to the positive square root of the number on the right side and solve for x.
6. 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.
Equation ( a )
Step 1: 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
Step 2: 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
Step 3: 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
Step 4: Take the square root of both sides.
4x² - 8x + 4 = 56
Square root (4x² - 8x + 4) = Square root (56)
(2x - 2)² = (2²*2*7)
(2x - 2)² = (2²)*(14)
(2x - 2) = 2*(14)
(To eliminate the 2’s on each side of the equation I am going to divide each side by 2. This simplifies the equation making it easier to solve)
(x - 1) = (1)*(14)
x - 1 = (1)*(14)
(Keep in mind that I am still taking the square root of each side. This will be consecutive throughout the equation)
Step 5: 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 = 1 + Sqrt(14)
x = 1 + 3.74166
x = 4.74166
Step 6: 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 = 1 - Sqrt(14)
x = 1 - 3.74166
x = -2.74166
the final solution is: x ∈ {-2.74166, 4.74166}
I am now going to go through the same exact six step process for equation ( c )
x² + 12x - 64 = 0
Step 1:
x² + 12x - 64 = 0
x² + 12x - 64 + 64 = 0 + 64
x² + 12x = 0 + 64
x² + 12x = 64
Step 2: Again 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
Step 3: The coefficient of the original x term is 12.
(12)² = 144
4x² + 48x = 256
4x² + 48x + 144 = 256 + 144
4x² + 48x + 144 = 400
Step 4: This is where we take the square root of both sides of the equation.
4x² + 48x + 144 = 400
Square root (4x² + 48x + 144) = Square root (400)
(2x + 12)² = (20²)
(2x + 12)² = ±20
2x + 12 = ±20
(Divide each side of the equation by 2 to simplify)
Step 5:
x + 6 = 10
x + 6 - 6 = 10 - 6
x = 4
Step 6:
x + 6 = -10
x + 6 - 6 = -10 - 6
x = -16
the final solution is: x ∈ {-16, 4}
Although quadratic equations require an extensive amount of work, we can luckily thank whomever in India for coming up with a system to break down all of the steps into an orderly manner so that the work is organized and easy to understand.
Project #2
For this project the directions ask to select five numbers, one being 0 (zero), two even numbers, and two odd numbers. The objective here is to use a formula that yields prime numbers and use the five randomly selected numbers, following the guidelines above, to solve the equation. Using the numbers I select I am going to substitute them in for x in the equation. After doing so, I will be determining which value of x will in fact yield a composite number. A composite number is “a positive integer that has a factor other than 1 and itself.“ (Bluman, 2011) With that in mind, it is important to remember that for a number to be prime, it can only have factors of 1 and itself. The difference between the two is that composite numbers has more than two factors whereas prime numbers only have 1.
Formula that yields prime numbers.
x² - x + 41
Select at least five numbers; 0 (zero), two even, and two odd.
My five numbers (x values):
Zero 0
Two even: 6, 8
Two odd: 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
71 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
Although by entering my selected numbers into the equation above I received all prime numbers for my solutions, but the expression x² - x + 41 does yield composite numbers as well. For example, if I were to have selected 45 as one of my numbers then we would take x = 45
x² - x + 41
= 45² - 45 + 41
= 2021
2021 is a composite number because it is a positive integer and it has two factors other than itself. Its factors are: 43 and 47.
Both projects for this assignment were very different in what exactly they were asking for, but with the use of the same similar equation set-up, I was able to formulate answers in which I hope are correct in what you were asking for.




Comments for Solving math equations - Mathematics in Our World - Projects - Page 397

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Mar 17, 2012
Solving math equations - Page 397
by: Staff



Answer:


For Project #1, work only equations (a) and (c)


It’s important to check your answers by substituting the solutions for x in the original equation.


Equation ( a )

x² - 2x - 13 = 0

your answer: 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




Equation ( c )

x² + 12x - 64 = 0

your answer: 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: Prime Numbers

Your solutions are correct



Thanks for writing.

Staff
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