#### notes

ax^{2} + bx + c, Divide the polynomial by a ( ∵a ≠ 0) to get `x^2+b/ax+c/a`.

Let us write the polynomial `x^2+b/ax+c/a` in the form of difference of two square numbers. Now we can obtain roots or solutions of equation `x^2+b/ax+c/a` which is equivalent to `ax^2 + bx + c = 0 .`

`ax^2 + bx + c = 0` .............(I)

`x^2+b/ax+c/a=0` ..... dividing both sides by a

`therefore x^2+b/ax+(b/(2a))^2-(b/(2a))^2+c/a=0`

`therefore (x+b/(2a))^2-b^2/(4a^2)+c/a=0`

`therefore(x+b/(2a))^2-(b^2-4ac)/(4a^2)=0` `therefore(x+b/(2a))^2=(b^2-4ac)/(4a^2)`

`therefore(x+b/(2a))=sqrt((b^2-4ac)/(4a^2)) or (x+b/(2a))=-sqrt((b^2-4ac)/(4a^2))`

`therefore x=-b/(2a)+sqrt((b^2-4ac)/(4a^2)) or x=-b/(2a)-sqrt((b^2-4ac)/(4a^2))`

`therefore x=(-b+sqrt(b^2-4ac))/(2a) or x=(-b-sqrt(b^2-4ac))/(2a)`

In short the solution is written as `x=(-b+-sqrt(b^2-4ac))/(2a)` and these values are denoted by `alpha,beta`.

`therefore alpha=(-b+sqrt(b^2-4ac))/(2a) , beta=(-b-sqrt(b^2-4ac))/(2a)` ............................(I)

The values of a, b, c from equation ax^{2} + bx + c = 0 are substituted in `(-b+-sqrt(b^2-4ac))/(2a)` and further simplified to obtain the roots of the equation. So `x=(-b+-sqrt(b^2-4ac))/(2a)` is the formula used to solve quadratic equation. Out of the two roots any one can be represented by α and the other by β.

That is, instead (I) we can write `alpha=(-b+sqrt(b^2-4ac))/(2a) , beta=(-b-sqrt(b^2-4ac))/(2a)` .......(II)