Slide 1 : QUADRATIC EQUATION
Quadratic Equation- Introduction:
In Our earlier classes we have studied about polynomials. The simplest of the polynomial are those which contain one term such as 8x, 3y2,, 4z3.These are called Monomials. A Polynomial such as 3x+2, x2-3 which contains two terms is called binomial. Polynomials containing three terms are called Trinomials.
X2+3x+2, 4y2+9y+2 are the examples for trinomials. A Polynomial of degree one is called a Linear Polynomial and that degree two is called a quadratic polynomial.In This chapter we are going to discuss Quadratic Polynomial,I mean quadratic equations.
Definition of Quadratic equations:
An equation in which one or more of the terms is squared but raised to no higher power.
Slide 2 : Zeros of a Quadratic Polynomial :
Zeros of a quadratic polynomial are nothing but the roots of the particular quadratic polynomial. I mean the solutions.
Let us consider the quadratic polynomial p (x) = x2+x+2
Now I substitute some real values of x and finding corresponding values of p (x).
x p (x)
----------------------------------------
0 -2
-1 -2
1 0
-2 0
+2 4
Thus by trial and error, we have found two values of x which are x=1 and x=2 for which the value of the quadratic equation x2+x+2 is zero. Each of these values is called zero of the given quadratic polynomial.
Slide 3 : I mean these two values are the roots of the given quadratic polynomial.
It is fact that every quadratic polynomial can have a most two zeros. I mean each and every quadratic polynomial would have two roots.
We have seen above that the second degree polynomials are commonly known as quadratic equation.
General equation of a quadratic equation : ax2+bx+c=0, here a, b , and c are real numbers and a is not equal to zero.
Example 1: Examine whether the equation
(i) (x-2)(x+3)+1=0
(ii) (x+4)(x-4)=x(x+2)+8
(iii) 6-x(x2+2)=0
are quadratic equation
Example 2: For the quadratic equation 2x2-5x-3=0,determine which of the following is (are) solutions?
(i) x=3 (ii) x=2 (iii) x=-1/2
Slide 4 : Exercise4.1:
Check whether the following equations are quadratic or not:
1. x2-6x-4=0 2. 3x2-7x-2=0
3. x3-6x2+2x-1=0 4.7x=2x2
5. x2+ 1 = 2 (x not = 0) 6.3x2-4=0
Solving a quadratic equation by Factorization:
We have already learnt how to factorize polynomial. In this section the method of factorization will be applied to solve quadratic equations.
In this method, first of all we will factorize the given quadratic equation and get two factors like x-a and x-b i.e (x-a)(x-b)=0
We are already familiar with the concept that
If the product of two numbers are zero, then at least one of them must be zero.
According to the above concept we can say either x-a= 0 or x-b=0
x=a x=b
Thus we get two values for x
Slide 5 : In this method we are going to use some algebraic identities
such as a2 – b2 = (a+b)(a-b)
(a + b)2 = a2+2ab+b2
(a – b)2 = a2 -2ab+b2
Example 3: Solve 64x2 -625=0
Example 4: Solve the quadratic equation : 16x2 -24x=0
Example 5: Solve: 25x2-30x+9=0
Example 6: Find the solutions of the quadratic equation x2+6x+5=0 check the solutions.
Example 7: Solve 6x2+x-15=0
Example 8: Solve 2x 1 3x+9
-------- + ---------- + ----------------------- = 0
x-3 2x-3 (x-3)(2x+3)
Exercise 4.2
Using Factorization and solve each of the following quadratic equations:
1. 9x2 -16 = 0 2. 64x2 -9 = 0
3. (x-2)2 – 25 = 0 4. (x+5)2 -36 = 0
5. ax2 -2abx = 0 5. 3x2 +10 = 11x