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  1. #include <stdio.h>
  2. #include <stdlib.h>
  3. #define VOID 0
  4.  
  5. double sqrt(double n) {
  6.  
  7.   double x = n, y = 1.0, e = 0.000001;
  8.  
  9.   while(x - y > e) {
  10.  
  11.     x = (x + y) / 2;
  12.  
  13.     y = n / x;
  14.  
  15.   }
  16.   return x;
  17. }
  18.  
  19. void NatureRootsQuadraticEquation(double a, double b, double c) {
  20.  
  21.     double S = -b/a,
  22.             P = c/a,
  23.             discriminant = b*b - 4*a*c,
  24.             x1, x2;
  25.     x1 = (-b + sqrt(discriminant)) / (2 * a);
  26.     x2 = (-b - sqrt(discriminant)) / (2 * a);
  27.     printf("S=%.2lf P=%.2lf\n",S,P);
  28.     printf("Delta = %.2lf\n\tx1=%lf\n\tx2=%lf\n", discriminant, x1, x2);
  29.  
  30.     if(discriminant >= 0) {
  31.  
  32.       if(S >= 0) {
  33.  
  34.          if(P > 0) {
  35.  
  36.             printf("%s","x1>0; x2>0");
  37.  
  38.          } else if(P < 0) {
  39.  
  40.             printf("%s","x1>0; x2<0; |x1| < |x2|");
  41.  
  42.          } else {
  43.  
  44.             printf("%s","x1=0; x2 >0");
  45.          }
  46.       //it means S < 0
  47.       } else {
  48.  
  49.         if(P > 0) {
  50.  
  51.            printf("%s","x1<0; x2<0");
  52.  
  53.         } else if(P < 0) {
  54.  
  55.            printf("%s","x1>0; x2<0");
  56.  
  57.         } else {
  58.  
  59.            printf("%s","x1=0; x2 <0");
  60.         }
  61.       }
  62.     } else {
  63.       printf("Imaginary");
  64.     }
  65.     printf("\n");
  66. }
  67.  
  68. int main(int argc, char const *argv[]) {
  69.   //declare three variable of double
  70.   double a, b, c;
  71.   a = 1;
  72.   b = 3;
  73.   c = -2;
  74.   //if the number of the arguments is greater than one
  75.   if(argc > 1){
  76.     //convert string to double
  77.     a = atof(argv[1]);
  78.     b = atof(argv[2]);
  79.     c = atof(argv[3]);
  80.   }
  81.  
  82.   //tests
  83.   if(b < 0 && c < 0) {
  84.        printf("%.3lfx^2 - %.3lfx - %.3lf = 0\n", a, -b, -c);
  85.   } else if(b < 0) {
  86.        printf("%.3lfx^2 - %.3lfx + %.3lf = 0\n", a, -b, c);
  87.   } else if(c < 0) {
  88.        printf("%.3lfx^2 + %.3lfx - %.3lf = 0\n", a, b, -c);
  89.   } else if(a > 0 && b > 0 && c > 0){
  90.        printf("%.3lfx^2 + %.3lfx + %.3lf = 0\n", a, b, c);
  91.   }
  92.   //solve
  93.   printf("%s\n", "Output:");
  94.   NatureRootsQuadraticEquation(a, b, c);
  95.  
  96.   return VOID;
  97. }
Success #stdin #stdout 0s 5304KB
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stdin
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Standard input is empty
stdout
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1.000x^2 + 3.000x - 2.000 = 0
Output:
S=-3.00 P=-2.00
Delta = 17.00
	x1=0.561553
	x2=-3.561553
x1>0; x2<0
https://ideone.com/jHvYz5
Nature Roots Quadratic Equation based on Viete's formulas
language:
C (gcc 8.3)
created:
1 year ago
visibility:
secret

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