C program for solving 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





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7 comments:

  1. #include
    #include
    #include
    main()
    {
    float a,b,c,d,r1,r2;
    printf("enter a ,b,c, values");
    scanf("%f%f%f",&a,&b,&c);
    d=(b*b)-(4*a*c);
    if(d>0)
    {
    r1=-b+sqrt(d)/2*a;
    r2=-b-sqrt(d)/2*a;
    printf("root1= %f,root2=%f",r1,r2);
    }
    else
    {
    printf("roots are imaginary");
    }
    .....................

    ReplyDelete
  2. Its a simple and best program.

    ReplyDelete
  3. U are really helpful .
    Thank u very much

    ReplyDelete
  4. Thank you very much for this easiest program

    ReplyDelete
  5. it gives call of nonfunction error

    ReplyDelete
  6. printf("%.3f%+.3fi",-b/(2*a),sqrt(-d)/(2*a));
    printf(", %.3f%+.3fi",-b/(2*a),-sqrt(-d)/(2*a));

    1) -b/(2*a)
    2) sqrt(-d) / (2*a)
    why these two expressions are separately evaluated?
    what is the meaning of these two expressions. Please anybody help me.

    ReplyDelete
  7. #include
    #include
    {void main()
    char arr[]="we are Bangladeshi";
    int len=strlen(arr[]);
    int i;
    int alpha
    for(i=0;arr[i]!='\0';i++)
    printf ("\n enter the line \n");
    printf("\n string \n");
    while(arr[i]='\0')
    if (arr[i]='32');
    {
    alpha_c++;
    }
    else {
    alpha_c
    }



    }

    ReplyDelete