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<math.h>

int 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");
    
    return 0;
}

Testing

Case 1:
Enter the values of a b c: 1 4 3
The real roots = -3.000000 -5.000000
Case 2:
Enter the values of a b c: 1 2 1
Roots are equal =-1.000000 -1.000000
Case 3:
Enter the values of a b c: 1 1 4
Roots are imaginary

Bhanu Priya

Updated on: 01-Sep-2023

102K+ Views

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