Write a program to solve a quadratic equation using the quadratic formula in c plus plus?

Answer 1

#include<stdio.h> #include<conio.h>

void main()

{

float a,b,c,z,d,x,y;

clrscr();

printf("Enter the value of a,b,c");

scanf("%f %f %f",&a,&b,&c);

d=((b*b)-(4*a*c));

z=sqrt(d);

x=(-b+z)/(2*a);

y=(-b+z)/(2*a);

printf("The Quadratic equation is x=%f and y=%f",x,y);

getch();

}

This answer does not think about imaginary roots. I am a beginner in C Programming. There might be flaws in my code. But, it does well.

the code is

#include<stdio.h>

#include<math.h>

main()

{

float a, b, c;

float dis, sqdis, real, imag, root1, root2;

printf("This program calculates two roots of quadratic equation of the form ax2+bx+c=0.\n");

printf("\n");

printf(" please type the coefficients a, b and c\n");

printf("\n");

printf("a = ");

scanf("%f", &a);

printf("\n");

printf("b = ");

scanf("%f", &b);

printf("\n");

printf("c = ");

scanf("%f", &c);

printf("\n");

dis = (b*b-4*a*c);

if(dis < 0)

{

sqdis = sqrt(-dis);

real = -b/(2*a);

imag = sqdis/(2*a);

printf(" The roots of the quadratic equations are \n x1\t=\t %f + %f i\n x2\t=\t %f -

%f i\n", real, imag, real, imag);

}

else

{

sqdis = sqrt(dis);

root1 = -b/(2*a)+sqdis/(2*a);

root2 = -b/(2*a)-sqdis/(2*a);

printf("The two roots of the quadratic equations are %f and %f.\n", root1, root2);

}

system("pause");

}

Answer 2

/*---------------------------------------------------------------------------

PROGRAM TO FIND THE ROOTS OF A QUADRATIC EQUATION

---------------------------------------------------------------------------*/

#include

#include

#include

int main()

{

int A,B,C;

float disc,deno,x1,x2;

printf("\n-------------------------------------------------------");

printf("\n\n PROGRAM TO FIND THE ROOTS OF A QUADRATIC EQUATION ");

printf("\n\n-------------------------------------------------------");

printf("\n\n\t ENTER THE VALUES OF A,B,C...");

scanf("%d,%d,%d",&A,&B,&C);

disc = (B*B)-(4*A*C);

deno = 2*A;

if(disc > 0)

{

printf("\n\t THE ROOTS ARE REAL ROOTS");

x1 = (-B/deno)+(sqrt(disc)/deno);

x2 = (-B/deno)-(sqrt(disc)/deno);

printf("\n\n\t THE ROOTS ARE...: %f and %f\n",x1,x2);

}

else if(disc == 0)

{

printf("\n\t THE ROOTS ARE REPEATED ROOTS");

x1 = -B/deno;

printf("\n\n\t THE ROOT IS...: %f\n",x1);

}

else

printf("\n\t THE ROOTS ARE IMAGINARY ROOTS\n");

printf("\n-------------------------------------------------------");

getch();

}

Answer 3

#include

#include

#include

void main()

{

int a,b,c,d;

float x1,x2,x,realp,imp;

clrscr();

printf("Enter the values of a,b & c:\n");

scanf("%d %d %d",&a,&b,&c);

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

if(d>0)

{

x1=-b+sqrt(d)/(2*a);

x2=-b-sqrt(d)/(2*a);

printf("\nThe Roots are real and not equal");

printf("\nThe roots are........x1=%f x2=%f",x1,x2);

}

else if(d==0)

{

x=-b/(2*a);

printf("\nThe roots are real and repeated");

printf("\nHence the repeated root x is %f",x);

}

else

{

realp=sqrt(abs(d))/(2*a);

imp=sqrt(abs(d))/(2*a);

printf("\nThe roots are complex or imaginary ");

printf("\n root1=%f+i%f",realp,imp);

printf("\n root2=%f-i%f",realp,imp");

}

getch();

}

Answer 4

#include
#include

using std::cin;
using std::cout;

int main()
{
cout << endl << "This program find real roots of a*x*x + b*x + c = 0";

double a = 0.0;
cout << endl << "Enter a: ";
cin >> a;

double b = 0.0;
cout << endl << "Enter b: ";
cin >> b;

double c = 0.0;
cout << endl << "Enter c: ";
cin >> c;

double det = b*b - 4*a*c;
double x1 = 0.0;
double x2 = 0.0;
if (det >= 0.0)
{
if (a == 0.0)
{
cout << endl << "It's a linear system with solution: ";
if (b != 0.0)
{
x1 = -c/b;
cout << x1;

}
else
{
cout << "System is not correct!";
return 1;

}

}
else
{
x1 = (-b + sqrt(det))/(2*a);
cout << endl << "Rirst root is: " << x1;

x2 = (-b - sqrt(det))/(2*a);
cout << endl << "Second root is: " << x2;

}

}
else
{
cout << endl << "There are no real roots!";

}

system("PAUSE");
return 0;

}

Answer 5

1. # include
2. # include
3. # include
4. void main()
5. {
6. float a, b, c, d, real, imag, r1, r2, n ;
7. int k ;
8. clrscr() ;
9. printf("Enter the values of A, B & C : ") ;
10. scanf("%f %f %f", &a, &b, &c) ;
11. if(a != 0)
12. {
13. d = b * b - 4 * a * c ;
14. if(d < 0)
15. k = 1 ;
16. if(d == 0)
17. k = 2 ;
18. if(d > 0)
19. k = 3;
20. switch(k)
21. {
22. case 1 :
23. printf("\nRoots are imaginary\n") ;
24. real = - b / (2 * a) ;
25. d = - d ;
26. n = pow((double) d, (double) 0.5) ;
27. imag = n / (2 * a) ;
28. printf("\nr1 = %7.2f + j%7.2f", real, imag) ;
29. printf("\nr2 = %7.2f - j%7.2f", real, imag) ;
30. break ;
31. case 2 :
32. printf("\nRoots are real and equal\n") ;
33. r1 = - b / (2 * a) ;
34. printf("\nr1 = r2 = %7.2f", r1) ;
35. break ;
36. case 3 :
37. printf("\nRoots are real and unequal\n") ;
38. r1 = (- b + sqrt((double) d)) / (2 * a) ;
39. r2 = (- b - sqrt((double) d)) / (2 * a) ;
40. printf("\nr1 = %7.2f",r1) ;
41. printf("\nr2 = %7.2f",r2) ;
42. break ;
43. }
44. }
45. else
46. printf("\nEquation is linear") ;
47. getch() ;
48. }

Answer 6

#include

void main()

{

float a,b,c,z,d,x,y;

printf("Enter the value of a,b,c");

scanf("%f %f %f",&a,&b,&c);

d=((b*b)-(4*a*c));

z=sqrt(d);

x=(-b+z)/(2*a);

y=(-b+z)/(2*a);

printf("The Quadratic equation is x=%f and y=%f",x,y);

}

Read more: http://wiki.answers.com/Write_a_program_in_c_programming_language_that_computes_the_roots_of_the_quadratic_equation_ax2_bx_c#ixzz1Uuc5ZW4a

Answer 7

The most straightforward way to do this is to use the quadratic equation.