QUADRATIC EQUATION : How to find solution (roots or zeroes) by Factorisation method ?

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 +x+1 (Quadratic).

In equation have a higher degree is 3 then it will be known as a cubic polynomial as shown +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.


The general form of the Quadratic equation is ax²+bx+c. where a is the coefficient of x², b is the coefficient of x & c is a constant term.

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 -


= 2x²-5x+3 = 0

= 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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