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Saturday, 3 December 2011

Write a c program to solve quadratic equation.

1. C program to calculate roots of a quadratic equation
2. Quadratic equation in c language
#include<stdio.h>
#include<math.h>
int main(){
  float a,b,c;
  float d,root1,root2;  
   printf("Enter a, b and c of quadratic equation: ");
  scanf("%f%f%f",&a,&b,&c);
   
  d = b * b - 4 * a * c;
  
  if(d < 0){
    printf("Roots are complex number.\n");
    printf("Roots of quadratic equation are: ");
    printf("%.3f%+.3fi",-b/(2*a),sqrt(-d)/(2*a));
    printf(", %.3f%+.3fi",-b/(2*a),-sqrt(-d)/(2*a));
  
    return 0; 
  }
  else if(d==0){
   printf("Both roots are equal.\n");
   root1 = -b /(2* a);
   printf("Root of quadratic equation is: %.3f ",root1);
   return 0;
  }
  else{
   printf("Roots are real numbers.\n");
  
   root1 = ( -b + sqrt(d)) / (2* a);
   root2 = ( -b - sqrt(d)) / (2* a);
   printf("Roots of quadratic equation are: %.3f , %.3f",root1,root2);
  }
  return 0;
}
Sample output:
Enter a, b and c of quadratic equation: 2 4 1
Roots are real numbers.
Roots of quadratic equation are: -0.293, -1.707
1. How to find a b and c in a quadratic equation
#include<stdio.h>
#include<math.h>
int main(){
  float a,b,c;
  float d,root1,root2;  
  printf("Enter quadratic equation in the format ax^2+bx+c: ");
  scanf("%fx^2%fx%f",&a,&b,&c);
   
  d = b * b - 4 * a * c;
  
  if(d < 0){
    printf("Roots are complex number.\n");
   
    return 0;
  }
 
   root1 = ( -b + sqrt(d)) / (2* a);
   root2 = ( -b - sqrt(d)) / (2* a);
   printf("Roots of quadratic equation are: %.3f , %.3f",root1,root2);
  return 0;
}
Sample output:
Enter quadratic equation in the format ax^2+bx+c: 2x^2+4x+-1
Roots of quadratic equation are: 0.000, -2.000

1 comment:

Average said...

#include
#include
#include
void main()
{
void quad(int a, int b ,int c);
int a,b,c;
float r;
clrscr();
printf(“Enter values of a,b,c of a quadratic equation:\n” );
scanf(“%d%d%d”,&a,&b,&c);
quad(a,b,c);
getch();
}
void quad(int a, int b ,int c)
{
float d,r1,r2,r3,r4,r;
d=(b*b)-(4*a*c);
switch(d)
{
case <0:
printf(“Value of Discriminant is negative” );
break;
case 0:
printf(“ Roots are real” );
r=-b/2*a;
printf(“ First and Second root of equation:%d,r” );
break;
default:
r1=-b+sqrt(d);
r2=2*a;
r=r1/r2;
r3=-b-sqrt(d);
r4=r3/r2;
break;
printf(“First root of equation: %d”,r);
printf(“/n Second root of equation: %d”,r4);
}}

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