C program to calculate the roots of a quadratic equation using two pointer parameters?

Answer 1

#include <stdio.h> #include <conio.h> #include <math.h> void main() { float a, b, c; float root1, root2,de,d; printf("enter the values of a, b and c"); scanf("%f%f%f", &a,&b,&c); de=(b*b)-4*(a*c); d=sqrt(de); root1=(-b+d) /(2.0*a); root2=(-b-d) /(2.0*a); printf(" roots of the equation are %f %f", root1,root2); getch(); }

Answer 2

Here is an example of a C program that calculates the roots of a quadratic equation using two pointer parameters:

#include <stdio.h>
#include <math.h>

void calculateRoots(float a, float b, float c, float* root1, float* root2) {
    float discriminant;

    discriminant = b*b - 4*a*c;

    if(discriminant > 0) {
        *root1 = (-b + sqrt(discriminant)) / (2*a);
        *root2 = (-b - sqrt(discriminant)) / (2*a);
        printf("Roots are real and different.\n");
    } else if(discriminant == 0) {
        *root1 = *root2 = -b / (2*a);
        printf("Roots are real and the same.\n");
    } else {
        printf("Roots are complex and different.\n");
    }
}

int main() {
    float a, b, c, root1, root2;

    printf("Enter coefficients a, b, and c: ");
    scanf("%f %f %f", &a, &b, &c);

    calculateRoots(a, b, c, &root1, &root2);

    printf("Root 1: %.2f\n", root1);
    printf("Root 2: %.2f\n", root2);

    return 0;
}

In this program, the calculateRoots function takes the coefficients a, b, and c of the quadratic equation as well as two pointer parameters root1 and root2 to store the roots. It calculates the roots using the quadratic formula and updates the values pointed to by the pointers. The main function prompts the user to enter the coefficients, calls the calculateRoots function, and then prints the roots.