C programming code for newton raphson method to solve quadratic equation?

Answer 1

yes

This is the codes for Newton-Raphson method for solving Quadratic equations

#include<stdio.h>

#include<conio.h>

#include<math.h>

#define F(x)(x*x*x)-(4*x)

#define FD(x)(3*x*x)-4

#define MAXIT 20

void main()

{

int count;

float x0,x1,fx,fdx;

clrscr();

printf("NEWTON-RAPHSON METHOD\n");

printf("---------------------\n");

printf("initial value:");

scanf("%f",&x0);

count=1;

begin:

fx=F(x0);

fdx=FD(x0);

x1=(x0-(fx/fdx));

if(fabs((x1-x0)/x1)<0.00001)

{

printf("The root is:%f\n",x1);

printf("Iteration is:%d\n",count);

}

else

{

x0=x1;

count=count+1;

if((count<MAXIT));

goto begin;

}

getch();

}

Answer 2

#include<stdio.h>

#include<conio.h>

#include<math.h>

#include<process.h>

void main()

{

float x0=0,x1=0,a=0,b=0,c=0,f0=0,e=0,df0,slope=.0001;

int i=0,n=0;

clrscr();

printf("\n Enter the interval");

printf("\n Enter the value of x0= ");

scanf("%f",&x0);

printf("\n Enter the error allowed in the root of equation ");

scanf("%f",&e);

printf("\n Enter the cofficients of the equation");

printf("\n Enter the value of a = ");

scanf("%f",&a);

printf("\n Enter the value of b= ");

scanf("%f",&b);

printf("\n Enter the value of c= ");

scanf("%f",&c);

printf("\n Enter the number of iterations= ");

scanf("%d",&n);

for(i=0;i<n;i++)

{

f0=a*x0*x0+b*x0+c;

df0=2*a*x0+b;

if((fabs(df0))<=slope)

{

printf("\n Slope too small so will become divergent from solution");

getch();

exit(0);

}

else

{

x1=x0-(f0/df0);

if(fabs((x1-x0)/x1)<e)

{

printf("\n Root of the equation is %f",x1);

printf("\n No of iterations for calculating the root of the equation is %d",i);

getch();

exit(0);

}

else

{

x0=x1;

}

}

}

printf("\n Solution does not converge in %d iterations",n);

getch();

}

Answer 3

/*programm for modified newton raphson method*/

#include<stdio.h>

#include<math.h>

#define epsilon 1e-6

main()

{

/*This is a programm based on modified

newton raphson method to

find out root of the equation x^3-5*x^2+7*x-3=0*/

double g,g1,v,v1,v2,x,dx;

int converged=0,i;

printf("plz enter the guess value\n");

scanf("%f",&g1);

i=1;

while (converged==0)

{

printf("\n iteration no=%d\n",i);

v=g1*g1*g1-5*g1*g1+7*g1-3;

printf("v=%lf\n",v);

v1=3*g1*g1-10*g1+7;

printf("v1=%lf\n",v1);

v2=6*g1-10;

printf("v2=%lf\n",v2);

x=(v*v1)/(v1*v1-v*v2);

printf("value=%lf\n",x);

g=g1-((v*v1)/(v1*v1-v*v2));

printf("new guess is=%lf\n",g);

dx=((g-g1)/g);

printf("error=%lf\n",dx);

g1=g;

if (fabs(dx)'less than'epsilon)

{converged=1;}

printf("the root of equation is=%lf

i=%d\n",g,i);

i=i+1;

}

}

Answer 4

Function f(x)

x2=x*x

f=x*x2-3*x2+2*x-1

return

end

function f1(x)

f1=3*x*x-6*x+2

return

end

5 write(*,*)'enter initial guess'

read(*,*) x0

write(*,*)'enter tolerrence'

read(*,*) eps

10 x1=x0-f(x0)/f1(x0)

if(abs((x1-x0)/x0).lt.eps) then

write(*,1) x1

1 format('solution=',f10.4)

stop

else

x0=x1

go to 10

endif

end