Colleen Cooper
Solving Quadratic Equations
MAT 126 Survey of Mathematical Methods
Instructor: Kussiy Alyass
October 1,, 2012

Solving Quadratic Equations
Using correct methods to solve quadratic equations can make math an interesting task. In the paper below I will square the coefficient of the x term, yield composite numbers, move a constant term and see if prime numbers occur. I will use the text and the correct formulas to create the proper solutions of the two projects that are required and solve for the equations... In project one, I will move the constant term to the right side of the equation and square the coefficient of the original x term and add it to both sides of the equation. Solve for project one: Part A

X2 – 2x – 13 = 0.
4x*4 + 8 = 52
4x*4 + 8 +16 = 52 + 16
4x*4 + 8 + 16 = 68
2x + 4 = 12 2x + 12 + -12
2x = 4 2x = -6
X = 2 x = -3
Part C.
X2 + 12x – 64 + 0
4xsquared + 12x = 64 – 4 squared
16 + 48 = 64 – 16
64 = 48 + 4x squared
In project two, I will substitute numbers for x to see if prime numbers occur and then try to find a number for x when substituted in the formula, yields a composite number. Project 2, Part A.
X2 – x + 41
8 x 8 + 41= 105 not a prime number
3 x 3 + 41 = 50 not a prime number
7 x 7 + 41 = 90 not a prime number
2 x 2 + 41 = 46 not a prime number
6 x 6 + 41 = 87 Prime number
In the work of this paper I was able to move the coefficient, yield composite numbers, move a constant term and see if prime numbers occur. I used the correct formulas and created the right solution for the two projects that I was required to do. Also, while doing these projects I learned about coefficients and quadratic equations and was able to understand math a little better. As far as applying this knowledge of math, I cannot come up with a situation where it will be used. I believe that one day it will be used during my life, but I just have not recognized when and where it will...

...Quadraticequation
In elementary algebra, a quadraticequation (from the Latin quadratus for "square") is any equation having the form
where x represents an unknown, and a, b, and c represent known numbers such that a is not equal to 0. If a = 0, then the equation is linear, not quadratic. The numbers a, b, and c are the coefficients of theequation, and may be distinguished by calling them, the quadratic coefficient, the linear coefficient and the constant or free term.
Solving the quadraticequation
A quadraticequation with real or complex coefficients has two solutions, called roots. These two solutions may or may not be distinct, and they may or may not be real.
Factoring by inspection
It may be possible to express a quadraticequation ax2 + bx + c = 0 as a product (px + q)(rx + s) = 0. In some cases, it is possible, by simple inspection, to determine values of p, q, r, and s that make the two forms equivalent to one another. If the quadraticequation is written in the second form, then the "Zero Factor Property" states that the quadraticequation is satisfied if px + q = 0 or rx + s = 0. Solving these two linear...

...Summer 2010-3 CLASS NOTES CHAPTER 1
Section 1.1: Linear Equations
Learning Objectives:
1. Solve a linear equation
2. Solve equations that lead to linear equations
3. Solve applied problems involving linear equations
Examples:
1. [pic]
[pic]
3. A total of $51,000 is to be invested, some in bonds and some in certificates of deposit (CDs). If the amount invested in bonds is to exceed that in CDs by $3,000, how much will be invested in each type of investment?
4. Shannon, who is paid time-and-a-half for hours worked in excess of 40 hours, had gross weekly wages of $608 for 56 hours worked. What is her regular hourly wage?
Answers: 1. [pic]
2. [pic]
3. $24,000 in CDs, $27,000 in bonds 4. $9.50/hour
Section 1.2: QuadraticEquations
Learning Objectives:
1. Solve a quadraticequation by (a) factoring, (b) completing the square, (c) the
quadratic formula
2. Solve applied problems involving quadraticequations
Examples:
1. Find the real solutions by factoring: [pic]
2. Find the real solutions by using the square root method: [pic]
3. Find the real solutions by completing the square: [pic]
4. Find the real solutions by using the quadratic formula: [pic]...

...Jaquavia Jacques
Ms. Cordell
1st period
December 9, 2014
Quadratics is used to help to determine what is on a graph. There are many formulas that are used to put points on a graph to create parabolas. Parabolas are “U” shaped figures on a graph. Parabolas are examples of quadratics on a graph. Parabolas can be positioned up or down, which means if the arrows are going up it has a minimum point, and if the arrows are going down that means it has a maximum point. When graphing using the vertex formula: or the roots formula: whether the “a” is positive or negative helps identify if the graph has a maximum or minimum. Below are some examples and a visual representation of what a parabola looks like at its minimum and maximum points.
Example1:
Type Of Formula
Equation
“a” Positive or Negative
Max or Min
Vertex Formula
Negative
Minimum
Roots Formula
Positive
Maximum
There are three positions you may see a parabola when it is on a graph. Each position of the parabola determines the Nature of Roots. A parabola eithers has two real roots, one real root, or no real roots. The way you determine if a parabola has two real roots if the parabola “cuts” or cross” the x- axis in two places.
The roots formula also shows when a parabola has two real roots, which is the reason it is called the roots formula: because you can identify the two real roots by looking at the formula. To identify the roots you will set the. equal to...

...A quadraticequation is an equation that has a second-degree term and no higher terms. A second-degree term is a variable raised to the second power, like x2. When you graph a quadraticequation, you get a parabola, and the solutions to the quadraticequation represent where the parabola crosses the x-axis.
A quadraticequation can be written in the form:quadraticequation,
where a, b, and c are numbers (a ≠0), and x is the variable. x is a solution (or a root) if it satisfies the equation ax2 + bx + c = 0.
Some examples of quadraticequations include:
3x2 + 9x - 2 = 0 6x2 + 11x = 7 4x2 = 13
--------------------------------------------------------------------------------
Solving a Quadratic Formula:
Some quadraticequations can be solved easily by factoring. Some simple-to-solve quadraticequations are:
x2 - 1 = 0
(x + 1)(x - 1) = 0
x = ±1
x2 - 5x + 6 = 0
(x - 2)(x - 3) = 0
x = 2, 3
Most second-degree equations are more difficult to solve, and cannot be solved by simple factoring. The quadratic formula is a general way of solving any quadraticequation:...

...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 quadraticequations. These particular methodologies or rather the identifiable method became founded and created in the country of India. For project one, the mathematician...

...While the ultimate goal is the same, to determine the value(s) that hold true for the equation, solvingquadraticequations requires much more than simply isolating the variable, as is required in solving linear equations. This piece will outline the different types of quadraticequations, strategies for solving each type, as well as other methods of solutions such as Completing the Square and using the Quadratic Formula. Knowledge of factoring perfect square trinomials and simplifying radical expression are needed for this piece. Let’s take a look!
Standard Form of a QuadraticEquation
ax2+ bx+c=0
Where a, b, and c are integers
and a≥1
I. To solve an equation in the form ax2+c=k, for some value k. This is the simplest quadraticequation to solve, because the middle term is missing.
Strategy: To isolate the square term and then take the square root of both sides.
Ex. 1) Isolate the square term, divide both sides by 2
Take the square root of both sides
2x2=40
2x22= 40 2
x2 =20
Remember there are two possible solutions
x2= 20
Simplify radical; Solutions
x= ± 20
x=± 25
(Please refer to previous instructional materials Simplifying Radical Expressions )
II. To...

...329
QuadraticEquations
Chapter-15
QuadraticEquations
Important Definitions and Related Concepts
1. QuadraticEquation
If p(x) is a quadratic polynomial, then p(x) = 0 is called
a quadraticequation. The general formula of a quadraticequation is ax 2 + bx + c = 0; where a, b, c are real
numbers and a 0. For example, x2 – 6x + 4 = 0 is a
quadraticequation.
2. Roots of a QuadraticEquation
Let p(x) = 0 be a quadraticequation, then the values of
x satisfying p(x) = 0 are called its roots or zeros.
For example, 25x2 – 30x + 9 = 0 is a quadraticequation.
3
And the value of x =
is the solution of the given
5
equation.
3
Since, if we put x =
in 25x2 – 30x + 9 = 0, we have,
5
2
3
3
LHS = 25 × – 30 ×
+ 9
5
5
= 9 – 18 + 9 = 0 = RHS
Finding the roots of a quadraticequation is known as
solving the quadraticequation.
5. Methods of SolvingQuadraticEquation
( i ) By Factorization
This can be understood by the examples given
below:
2
Ex. 1: Solve: 25 x 30 x 9 0
Soln: 25x 2 30x 9 0 is equivalent to
5x 2 25x 3 32
0
5 x 32 0
3 3
3
,
or...

...QUADRATICEQUATIONSQuadraticequations Any equation of the form ax2 + bx + c=0, where a,b,c are real numbers, a 0 is a quadraticequation.
For example, 2x2 -3x+1=0 is quadraticequation in variable x.
SOLVING A QUADRATICEQUATION
1.Factorisation
A real number a is said to be a root of thequadraticequation ax2 + bx + c=0, if aa2+ba+c=0. If we can factorise ax2 + bx + c=0, a 0, into a product of linear factors, then the roots of the quadraticequation ax2 + bx + c=0 can be found by equating each factor to zero.
Example – Find the roots of the equation 2x2 -5x +3=0, by factorisation.
Solution:
2x2 -5x +3=0 2x2 -2x-3x+3=0 2x(x-1)-3(x-1)=0 i.e., (2x-3)(x-1)=0 Either 2x-3=0 or x-1=0. So,the roots of the given equation are x=3/2 and x=1.
2. Completing the square
To complete the square means to convert a quadratic to its standard form. We want to convert ax2+bx+c = 0 to a statement of the form a(x h)2 + k = 0.
To do this, we would perform the following steps:
1) Group together the ax2 and bx terms in parentheses and factor out the coefficient a.
2) In the parentheses, add and subtract (b/2a)2, which is half of the x coefficient, squared.
3) Remove the term - (b/2a)2 from the...