Question

write a program to solve quadratic equation with the help of C language syntax

Answer 1

Program 1: calculate roots of a quadratic equation
#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;
}


Result
Enter a, b and c of quadratic equation: 2 4 1
Roots are real numbers.
Roots of quadratic equation are: -0.293, -1.707







Program 2: 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;
}


Result