Quadratic equations Any equation of the form ax2 + bx + c=0, where a,b,c are real numbers, a 0 is a quadratic equation.

For example, 2x2 -3x+1=0 is quadratic equation in variable x.

SOLVING A QUADRATIC EQUATION

1.Factorisation
A real number a is said to be a root of the quadratic equation 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 quadratic equation 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 parentheses. Don't forget to multiply the term by a, when removing from parentheses.

4) Factor the trinomial in parentheses to its perfect square form, (x + b/2a)2.

5) Transpose (or shift) all other terms to the other side the equation and divide each side by the constant a.

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6) Take the square root of each side of the equation.

7) Transpose the term -b/2a to the other side of the equation, isolating x.

Example-

The Quadratic Formula The method of completing the square can often involve some very complicated calculations involving fractions. To make calculations simpler, a general formula for solving quadratic equations, known as the quadratic formula, was derived. To solve quadratic equations of the form ax2 + bx + c = 0,...

...the nature of the roots of quadratics and cubic functions.
Part One
Case One
For Case One, the discriminant of the quadratic will always be equal to zero. This will result in the parabola cutting the axis once, or twice in the same place, creating a distinct root or two of the same root.
For PROOF 1, the equation y=a(x-b)2 is used.
PROOF 1
y = 3 (x – 2)2
= y = 3 (x2 – 4x + 4)
= y = 3x2 – 12x + 12
^ = b2 – 4ac
= (-12)2 – 4 x 3 x 12
= 144 – 144
= 0
The discriminant is equal to zero. The parabola touches the x-axis at (2, 0) and as a in the equation is a positive value, the parabola curves upwards.
For PROOF 2, the equation y=-a(x-b)2 is used.
PROOF 2
y = -3(x - 2)2
= y = -3 (x2- 4x + 4)
= y = -3x2 + 12x – 12
^ = b2 – 4ac
= 122 – 4 x (-3) x (-12)
= 144 – 144
= 0
The discriminant of the parabola is equal to zero. The parabola touches the x-axis at (2,0) and as a is a negative value, the parabola curves downwards.
For PROOF 3, the equation y=a(x-A)2 is used.
y = a (x – A)2
= y = a (x2 – 2Ax + A2)
= y = ax2 – 2aAx + aA2
^ = b2 – 4ac
= (- 2aA)2 – 4 x (a) x (aA2)
= 4a2A2 - 4a2A2
= 0
The discriminant of the parabola is equal to zero. The parabola touches the x- axis at (A, 0) and whether the parabola curves up or down is dependent on the whether A is a positive or negative value.
Case Two
In Case Two, the discriminant of the...

...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...

...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...

...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 equations provides the roots...

...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 Solving QuadraticEquation
( 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 simply x
as the
5 5
5
required...

...• A quadratic function (f) is a function that has the form as f(x) = ax2 + bx + c where a, b and c are real numbers and a not equal to zero (or a ≠ 0). • The graph of the quadratic function is called a parabola. It is a "U" or “n” shaped curve that may open up or down depending on the sign of coefficient a. Any equation that has 2 as the largest exponent of x is a quadratic function.
☺Forms of Quadratic functions:
*Quadratic functions can be expressed in 3 forms: 1. General form: f (x) = ax2 + bx + c 2. Vertex form: f (x)= a(x - h)2 + k (where h and k are the x and y coordinates of the vertex) 3. Factored form: f(x)= a(x - r1) (x - r2)
1. General form
• Form : f(x) = ax2+ bx+ c • General form is always written with the x2 term first, followed by the x term, and the constant term last. a, b, and c are called the coefficients of the equation. It is possible for the b and/or c coefficient to equal zero. Examples of some quadratic functions in standard form are: a. f(x) = 2x2 + 3x – 4 (where a = 2, b = 3, c = -4) b. f(x) =x2 – 4 (where a = 1, b = 0, c = -4) c. f(x)= x2 ( where a = 1, b and c = 0) d. f(x)= x2 – 8x (where a = ½, b = -8, c = 0).
2. Vertex Form
• Form: f(x) = a(x - h)2 + k where the point (h, k) is the vertex of the parabola. • Vertex form or graphing form of a parabola. • Examples: a. 2(x - 2)2 + 5 (where a = 2, h = 2, and k = 5)...

...Colleen Cooper
Solving QuadraticEquations
MAT 126 Survey of Mathematical Methods
Instructor: Kussiy Alyass
October 1,, 2012
Solving QuadraticEquations
Using correct methods to solve quadraticequations 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 +...