Showing posts with label Algebra Quadratic Equations. Show all posts
Showing posts with label Algebra Quadratic Equations. Show all posts

Monday, August 13, 2012

Introduction for Algebra Quadratic Equations

Introduction for Algebra Quadratic Equations:

Algebra :It is a branch of mathematics that includes terms,equations, polynomials and algebraic structures.The most basic form of algebra is elementary algebra.The terms that are related with algebra are
*Variables these are alphabetical characters that are used for assigning the value.generally we use x,y,z as variables.
*Terms:It is a single number or a variable and numbers multiplied together.Example 2x-5=3 here 2x,5,3 are called terms
*Constant :It is fixed value and does not contain any variable.Example the constant term in 3x+2 is 2.
In algebra polynomials is an expression that contains one or more variables and constants using operations(addition,subtraction and multiplication).Quadratic equation is an example of polynomials.

Quadratic Equation:
   It is an equation that contains a second Degree term. The quadratic equation is of the form ax^2+bx+c=0 where a, b, c are the coefficients (numbers) and X is the variable .The Variable X is the solution (or we can say it is a root) if it satisfies the given equation ax^2+bx+c=0.
Some Examples of quadratic equation are given as
x2+3x+4=0        3x2+9x-2=0            3x2
=12
Quadratic equations can be solved by:
  • Using Simple Factoring
  • Using Quadratic Formula
  • Using the square root method.
  • Graphing.
  • Completing the square method
Solving Quadratic Equations by Method of Factorization:

Step 1: Write the equation  in standard form, ax^2+bx+c=0,where a is not equal to zero, if needed.

Example: X2+7x-8=0

Step2: Find the product ac .Determine all possible factors of the first and third term ac= -8 all possible factors of 8 are (2 * 4) and (1 * 8).

Step3: Choose the factors that will add to give the middle term. Addition and subtraction sign tell you whether the term should be positive or negative. Since ac = - 8 we will need a positive and negative factor. Since the middle term, is positive, the positive factor of 8 will need to be larger than the negative factor. For instance, (-8*1) will give -8 but result is -7.So we take 1 as negative factor and 8 as positive.

Step 4: The final factors will be in the form (x+ ) (x+ ) 
This trinomial will have the factors (x-1) (x+8). 

Step5. 
Use the Zero product Principle 
To solve for X set each factor= 0
x-1= 0 that given x=1 and x+8= gives x= - 8

Quadratic Formula to Solve Quadratic Equation in Algebra

Quadratic formula is given as
      The General Quadratic Formula to find the roots of a quadratic equation is,
                        x = `((-b+-sqrt(b^2-4ac))/(2a))`
Where,
            a = coefficient of x2
            b= coefficient of x
            c= constant
            b2-4ac = discriminant

This formula generally gives 2 solutions. The Plus and minus sign shows that there are 2 possible 
solutions.
Deriving the Quadratic formula:
It is obtained by solving the general quadratic equation
  aX2+bX+c=0
Divide each side of equation by a
  X2+(b/a)X+c/a=0
Subtract c/a from each side of the equation
 X2+(b/a)X=-c/a
To complete the square add (b/2a)2 to each side of the equation
  X2 + (b/a)X + (b/2a)2=-c/a + (b/2a)2
  (X+b/2a)2= -c/a+ (b2/4a2)
Taking LCD for the right side of equation
(X+b/2a)2= (-4ac/4a2)+ (b2/4a2)
(X + b/2a)2= (-4ac +b2)/4a2
Take the square root of each side of the equation.
  X+ (b/2a) =  ±√(b2-4ac) / 2a
   X = `((-b+-sqrt(b^2-4ac))/(2a))`

When a quadratic equation is in standard form ax^2+bx+c=0 , the expression, b^2-4ac, that is found under the square root part of the quadratic formula is called the discriminant.
Discriminant formula 
    
 d=b2-4ac 
Kinds of solutions for ax2+bc+c=0
a)If Discriminant d is greater than zero(means it is positive) b2-4ac>0
 :The equation has two solutions both of which are real numbers. Or we can say it has 2 distinct real solutions
b)If Discriminant is zero  b2-4ac=0 : The equation has only one real solution
X=-b/2a
c) If Discriminant Is less than zero( Discriminant is negative)b2-4ac<0 span="span">  : No solutions are defined.
There are two distinct (non-real) complex roots, which are complex conjugates of each other:
  (-b + i √4a c-b2)/2a       and       (-b- i √4ac-b2)/ 2a
where i =`sqrt(-1)` 

Nature of Roots of a Quadratic Equation

The quantity D = b2- 4ac is known as the discriminant of the quadratic equation whose roots are given by:
        α =  `(-b + sqrt(b^2-4ac))/(2a)`     and
       β = `(-b- sqrt(b^2-4ac))/(2a)` 
The nature of roots is as given below:
  1. The roots are real and distinct iff D>0.
  2. The roots are real and equal iff D=0.
  3. The roots are complex with non-zero imaginary part iff D<0 .="." span="span">
  4. The roots are rational iff   a,b,c are rational and D is a perfect square.
  5. The roots are of the form p+sqrt(q) (p,q belongs to Q) iff a,b,c are rational and d is not a perfect square.
  6. If a=1, b,c belongs to I and the roots are rational number, then these roots must be integers.
  7. If a quadratic equation in x has more than two roots, than it is an identity in x that is a = b = c = 0. 

Square Root Method to Solve a Quadratic Equation in Algebra

Steps to Solve Quadratic equation Using Square Root method
Step 1: Write the quadratic equation in the form A2=B if needed.
 Example X2 = 16 

Step 2: Apply the square root method. 
 If A and B are algebraic expressions such that A2 = B ,then
 A is equal to Square root B or A is equal to - Square root B
 Also it is written as A = plus minus root B
X2=16
X=√16  or X= -√16
There are 2 solutions
Step 3: Solve for the linear equation(s) set up in step 2.
  X=√16  that means  X=4
 X=-√16  that means X=-4
There are two solutions to above quadratic equation that is given as x = 4 and x = -4.