Algebra/Chapter 23/Polynomial Roots

Theory of equations

A Fundamental subject in Mathematics, which deals with the methods of finding roots of a given mathematical expression.Suppose f(x) is a function in x and m is a constant in the range of the function.Now if :-

         f(x)=m 

,there are only a finite number of values of x which can satisfy the given expression. Of course this is not true for trigonometric equations which yield an infinite number of roots but usually only the principal values of angles are taken.

Solution to a Quadratic equation

Theoretically a polynomial equation of the nth degree can be resolved into n factors.Let us consider the polynomial equation:-

$F(x)=x^{n}+ax^{n-1}+bx^{n-2}.................$  where a,b are constants..Let F(x)=0 for some value of x.

let us assume F(x)=(x-m)(x-n)(x-o)(x-p).................. where m,n are called the roots of the equation F(x). Now if we multiply the factors we obtain:- $F(x)=x^{n}-(m+n+o......)x^{n-1}+(mn+no+.........)x^{n-2}................+mnop......$ Comparing with the polymomial,

 Sum of roots=-a
 Sum of roots taken two at a time=b
 ..............................
 ..............................
 Sum of roots taken k at a time=$(-1)^{k}$(co-efficient of the (k+1)st term)

This identity was discovered by Vieta and is named after him. Hence we have n equations to find n roots. Now consider an equation of degree 2 called the Quadratic equation.Let m,n be the roots of the equation. Let the equation be :-

    $ax^{2}+bx+c=0$  where a,b,c are constants

This can also be written as:- $x^{2}+{\frac {bx}{a}}+{\frac {c}{a}}=0$ dividing by a

From the above treatise we know that:-

$m+n={\frac {-b}{a}}$
$mn={\frac {c}{a}}$
also 
$(m-n)^{2}=(m+n)^{2}-4mn$
Hence:-
$m-n=((m+n)^{2}-4mn)^{1/2}$
$m-n=({\frac {b^{2}}{a^{2}}}-{\frac {4c}{a}})^{1/2}={\frac {(b^{2}-4ac)^{1/2}}{a}}$
Solving the above simultaneous equations,we get
$m={\frac {-b+(b^{2}-4ac)^{1/2}}{2a}}$ and $n={\frac {-b-(b^{2}-4ac)^{1/2}}{2a}}$