We generally observe some equation or expression in Mathematics like x+1(Linear), x²+x+1 (Quadratic), x³+x²+x+1 (Cubic) where x is a variable.
These equations may be classified according to their degree of the polynomial. First, we need to understand what is the degree of a polynomial.
Degree of a polynomial is a higher degree in any given expression represent by variables power.
Types of Equations
If in equation or expression higher degrees is 1 so it will be linear equation as shown x+1(Linear).
If the expression has a higher degree 2 then it will be a quadratic equation as shown x²+x+1 (Quadratic).
In equation have a higher degree is 3 then it will be known as a cubic polynomial as shown x³+x²+x+1 (Cubic).
If the expression has a higher degree 2 then it will be a quadratic equation as shown x²+x+1 (Quadratic).
In equation have a higher degree is 3 then it will be known as a cubic polynomial as shown x³+x²+x+1 (Cubic).
QUADRATIC EQUATION
An equation in which the degree of a polynomial is 2 then the expression is known as a quadratic equation.
Examples :- 2x²-5x+3, 6x²-x-2, 2x²+x-6, x²-3x-10.
If we compare the given example & general form of Quadratic Equation so we get below results-
2x²-5x+3 where a=2, b=-5 ,c=3
6x²-x-2 where a=6, b=-1, c=-2
2x²+x-6 where a=2, b=1, c=-6
x²-3x-10 where a=1, b=-3, c=-10
So how is
Quadratic Equation its variable & coefficient, I think you understand very
well.
Now we discuss how Factorisation method is used to find the
solution (roots or zeroes) of any Quadratic Equation.
FACTORISATION METHOD:-
It is a method
which is used to find the value of a variable or to find the solution (roots or
zeroes) for Quadratic equation.
We do factors of terms in Quadratic equation.
Yes, a question arising...How do we factor & find its solution?
Let's
discuss with an example-
Example:-
Find the zeroes of quadratic equation 2x²-5x+3.
Solve:-
We are taking a Quadratic equation 2x²-5x+3 & split its middle
term -5x. For split, it, first
multiply 2 (coefficient of x²) &
3 (constant term).
When
multiplying 2 & 3 we get 6
(2×3=6).
Now this 6 taken as reference for splitting.
Now look middle term -5x we need to split it.
We may split -5x in different ways (-1x - 4x = -5x) ,(-2x - 3x = -5x), (-6x + x = -5x) etc.
But remember when we multiply above both numbers (shown from red text) then we should get +6, like (-2 × - 3 = +6) & when we subtract or add both number then we get -5x ,like (-2x - 3x = -5x).
Condition satisfying by above-highlighting portion, so we use it for splitting.
Now find
its solution (roots or zeroes) & take equal to zero -
= 2x²-2x-3x+3 = 0 {Now, will take common from underline terms }
= 2x(x-1) - 3(x-1) = 0 {Taking common underline terms}
= (x-1)(2x-3) = 0 {Now, we will take underlined terms equal to zero}
= (x-1) = 0 , (2x-3) = 0
= x-1 = 0 , 2x-3 = 0
= x=1 , x = 3/2
Finally we get our answer (roots or zeroes) x=1 & x = 3/2 for Quadratic Equation 2x²-5x+3.
You may solve other examples from this method.
In some cases factorisation method does not work so we use another method. We will discuss more method in another post.
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