How to write a C program to find the roots of a quadratic equation?

Problem

Applying the software development method to solve any problem in C Language

Solution

• Find roots of a quadratic equation, ax2+bx+c.
• There will be 2 roots for given quadratic equation.

Analysis

Input − a,b,c values

Output − r1, r2 values

Procedure

$r_{1}=\frac{-b+\sqrt{b^2-4ac}}{2a}$

$r_{2}=\frac{-b-\sqrt{b^2-4ac}}{2a}$

Design (Algorithm)

• Start
• Read a, b, c values
• Compute d = b2 4ac
• if d > 0 then
  • r1 = b+ sqrt (d)/(2*a)
  • r2 = b sqrt(d)/(2*a)
• Otherwise if d = 0 then
  • compute r1 = -b/2a, r2=-b/2a
  • print r1,r2 values
• Otherwise if d < 0 then print roots are imaginary
• Stop

Implementation Code

# include<stdio.h>
# include<conio.h>
# include<math.h>
main (){
   float a,b,c,r1,r2,d;
   printf (“enter the values of a b c”);
   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 (“The real roots = %f %f”, r1, r2);
   }
   else if (d= =0){
      r1 = -b/(2*a);
      r2 = -b/(2*a);
      printf (“roots are equal =%f %f”, r1, r2);
   }
   else
      printf(“Roots are imaginary”);
   getch ();
}

Testing

Case 1: enter the values of a b c: 1 4 3
   r1 = -1
   r2 = -3
Case 2: enter the values of a b c: 1 2 1
   r1 = -1
   r2 = -1
Case 3: enter the values of a b c: 1 1 4
Roots are imaginary