# Ascertaining Mathematics with Equations

Topics: Prime number, Polynomial, 1 Pages: 7 (1612 words) Published: February 6, 2013
Ascertaining Mathematics With Equations
Jesse J. Oliver Jr.
Mathematics 126: Survey of Mathematical Methods
Professor Matthew Fife
Thursday, January 24, 2013

Ascertaining Mathematics With Equations
The abstract science of a number, quantity and space that can be studied in its very own right or as it may be applied to other disciplines and subject matters in several aspects, one considers to be that of mathematics. The problem of testing a given number for “primality” has been known to be proven by Euclid in ancient Greece that there are in fact infinitely many primes. In such relations, a mathematician will describe in example two projects that use prime numbers, composite numbers and the quadratic formula to solve equations. From the projects section on page 397 of Mathematics in Our World, for Project One, the mathematician will work only equations ( a ) and ( c ), but complete each of the six steps (a-f) and for Project Two, the mathematician will select a minimal of five numbers including zero (0) as one number and the other four are to be two even and two odd numbers.

PROJECT ONE
Basically, the foundation of Project One originates from a thought provoking methodology for finding solutions to quadratic equations. These particular methodologies or rather the identifiable method became founded and created in the country of India. For project one, the mathematician is to solve equations ( a ) and ( c ) and show the steps used to complete the equation. Equation ( a ) states that x^2 - 2x - 13 = 0 . Equation ( c ) states that x^2 + 12x - 64 = 0 . To solve each equation, both equation ( a ) and equation ( c ), the mathematician is to move the constant term to the right side of the equation to begin with. Following the initial step to transfer the constant term to the right side of the equation, the mathematician is to multiply each term in the equation by four times the coefficient of the x squared term. Next, square the coefficient of the original x term and add it to both sides of the equation. Then take the square root of both sides. Organize the left side of the equation so that it is equal to the positive square root of the number on the right side and of course solve for x. To draw a solution, the mathematician will then organize the left side of the equation so that it will be equal to the negative square root of the number on the right side of the equation and solve for x.

To solve Equation ( a ), the mathematician will derive at a solution by these steps:

Step 1: Move the constant term to the right side of the equation.

x^2 - 2x - 13 = 0
x^2 - 2x - 13 + 13 = 0 + 13
x^2 - 2x + 0 = 0 + 13
x^2 - 2x = 0 + 13
x^2 - 2x = 13

Step 2: Multiply each term in the equation by four times the coefficient of the x-squared term.

The coefficient of the x^2 term is 1.

x^2 - 2x = 13
(4 • 1) • (x^2 - 2x = 13)
(4) • (x^2 - 2x = 13)
(4) • x^2 + (4) • (-2x) = (4) • (13)
4x^2 - 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
4x^2 - 8x = 52
4x^2 - 8x + 4 = 52 + 4
4x^2 - 8x + 4 = 56

Step 4: Take the Square Root of both sides.

4x^2 - 8x + 4 = 56
Square Root (4x^2 - 8x + 4) = Square Root (56)
(2x - 2)^2 = (2^2 • 2 • 7)
(2x - 2)^2 = (2^2) • (14)
(2x - 2) = 2 • (14)

(To eliminate the 2’s on each side of the equation, the mathematician will divide each side by 2. This step simplifies the equation making it easier for the mathematician to solve)

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

(Be sure to remember that the mathematician will be 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 = + Square Root (14)
x - 1 + 1 = Square Root (14) + 1
x = 1 +...

Please join StudyMode to read the full document