This document provides information about quadratic equations including: 1) It defines a quadratic equation as a polynomial equation of the second degree in the general form of ax2 + bx + c = 0, where a ≠ 0. 2) It discusses the importance of quadratic equations, noting that the term "quadratic" comes from the variable being squared (x2) and that a quadratic equation is a trinomial expression with three terms. 3) It presents the method of factorization to solve quadratic equations by factorizing the left side of the equation into two binomial expressions and setting each factor equal to zero to solve for the solutions.

1. JSS MAHAVIDYAPEETA MYSURU 4
JSS INSTITUTE OF EDUCATION
Sakleshpur 573134
ANUSHA RANI A
B.ED 2nd SEMESTER
REG.NO: ED190306
JSS INSTIUTUTE OF EDUCATION
SAKLESHPUR

4. INTRODUCTION
CONTENT
Definition of quadratic equation
Examples
Hidden quadratic equation
Methods of solution
1. Factorization

5. DEFINITION
 In mathematics, a quadratic equation is a polynomial equation of
the second degree.
 The general form is ax2 + bx + c=0
Where x represents a variable or an unknown and a, b and c are
constants with a≠0.
If a=0??
Then the equation is a linear
equation

6. IMPORTANCE
 The general form is ax2 + bx + c = 0
 The expression on the right hand side ax2+ bx+ c is called as a
quadratic expression.
 The name quadratic comes from quad ,meaning square ,because
the variable gets squared like x2.
A quadratic equation is a trinomial expression.
Why???
Because in standard form it adds three terms
ax2 , bx and c.

7. More examples of quadratic equation
 2x2+ 5x+ c=0
In this one a=2,b=5 c=3.
 x2_ 3x=0
This one is little more tricky; Where is a?
In fact a=1, as we don't usually write “1x2” b= -3 and where is c? well, c=0 is
not shown.
 5x- 3=0
Oops! This one is not be quadratic equation because it is missing x2 (in other
Words if a=0 and that means it cant be quadratic).

8. METHODS OF SOLUTION
The methods of solving quadratic equation is :
• Factorization

9. Ex: Solve x2 + 7x + 6 = 0
Quadratic equation
 factor the left hand side (LHS)
x2 + 7x + 6 = (x + 6 )(x + 1)
 x2 + 7x + 6 = (x + 6)(x + 1) = 0
Now the equation as given is of the form ab = 0
 set each factor equal to 0 and solve
x + 6 = 0 x + 1 = 0
Solution: x = - 6 and – 1  x = {-6, -
1}

10. Method 1: Factorization
Q : Factorize ax2+ bx+ c
If ax2+ bx+ c = (rx + p)(sx + p) = 0
Then the solutions of the equation are
x1 = -p/r x2 = -q/s
Example:
x2 -2x -15=0
(x - 5) (x + 3)=0
x =5 x=-3
-15 x2
-5 +3 2
2

11. Any questions . . .