Best Answer

#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");

}

More answers

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

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();

}

#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();

}

#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;

}

- # include
- # include
- # include
- void main()
- {
- float a, b, c, d, real, imag, r1, r2, n ;
- int k ;
- clrscr() ;
- printf("Enter the values of A, B & C : ") ;
- scanf("%f %f %f", &a, &b, &c) ;
- if(a != 0)
- {
- d = b * b - 4 * a * c ;
- if(d < 0)
- k = 1 ;
- if(d == 0)
- k = 2 ;
- if(d > 0)
- k = 3;
- switch(k)
- {
- case 1 :
- printf("\nRoots are imaginary\n") ;
- real = - b / (2 * a) ;
- d = - d ;
- n = pow((double) d, (double) 0.5) ;
- imag = n / (2 * a) ;
- printf("\nr1 = %7.2f + j%7.2f", real, imag) ;
- printf("\nr2 = %7.2f - j%7.2f", real, imag) ;
- break ;
- case 2 :
- printf("\nRoots are real and equal\n") ;
- r1 = - b / (2 * a) ;
- printf("\nr1 = r2 = %7.2f", r1) ;
- break ;
- case 3 :
- printf("\nRoots are real and unequal\n") ;
- r1 = (- b + sqrt((double) d)) / (2 * a) ;
- r2 = (- b - sqrt((double) d)) / (2 * a) ;
- printf("\nr1 = %7.2f",r1) ;
- printf("\nr2 = %7.2f",r2) ;
- break ;
- }
- }
- else
- printf("\nEquation is linear") ;
- getch() ;
- }

#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);

}

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

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

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You don't need a flow chart for that; just use the quadratic formula directly; most programming languages have a square root function or method. You would only need to do this in many small steps if you use Assembly programming. The formulae would be something like this: x1 = (-b + sqrt(b^2 - 4*a*c)) / (2 * a) and x2 = (-b - sqrt(b^2 - 4*a*c)) / (2 * a) where a, b, and c are the coefficients of the quadratic equation in standard form, and x1 and x2 are the solutions you want.

A c program is also known as a computer program. A singular matrix has no inverse. An equation to determine this would be a/c=f. <<>> The determinant of a singular matix is zero.

/*Hello!! I'm aditya From Bangalore I have the solution of this program*/ #include <stdio.h> #include <math.h> main() { float a,b,c,x1,x2,delta=0; printf("enter the value of a,b,c\n"); scanf("f%f",&a,&b,&c); delta=((b*b)-(4ac)); if(a=0) { printf("the variables cannot form a quadratic equation\n"); } else { x1=(-b)+(sqrt(((b*b)-(4ac))/2a)) x2=(-b)-(sqrt(((b*b)-(4ac))/2a)) } if(delta=0) { printf("the roots are real and equal"); } if(delta>0) { printf("the roots are real and distinct"); } if(delta<0) { printf("the roots are imaginary"); x1=(-b)+(sqrt(((b*b)-((4ac))/(float)(2a)) x2=(-b)-(sqrt(((b*b)-((4ac))/(float)(2a)) } printf("the roots of the equation are %2.2f\n",x1,x2,delta); } /*the program is written by using simple-if and if-else constructs*/

(Uses Square Root Function) PRINT "Ax^2 + Bx + C = 0" INPUT "A = ", A INPUT "B = ", B INPUT "C = ", C D = B * B - 4 * A * C IF D > 0 THEN DS = SQR(D) PRINT "REAL ROOTS:", (-B - D) / (2 * A), (-B + D) / (2 * A) ELSE IF D = 0 THEN PRINT "DUPLICATE ROOT:", (-B) / (2 * A) ELSE DS = SQR(-D) PRINT "COMPLEX CONJUGATE ROOTS:", (-B / (2 * A)); "+/-"; DS / (2 * A); "i" END IF END IF

How to write a program for secant method by mathematica

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The easiest way to write a generic algorithm is to simply use the quadratic formula. If it is a computer program, ask the user for the coefficients a, b, and c of the generic equation ax2 + bx + c = 0, then just replace them in the quadratic formula.

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