#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");
}
/*---------------------------------------------------------------------------
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
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.
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
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.
Write an algorithm to find the root of quadratic equation
First, write the equation in standard form, i.e., put zero on the right. Then, depending on the case, you may have the following options:Factor the polynomialComplete the squareUse the quadratic formula
to solve ax2 + bx + c use the quadratic formula: (-b +/-(b2 - 4ac))/2a. Programming this should be a doddle.
readuse the answer
2000X=Y2KoverZzz?
Write your program and if you are having a problem post it here with a description of the problem you are having. What you are asking is for someone to do your homework for you.
Write the quadratic equation in the form ax2 + bx + c = 0 then the roots (solutions) of the equation are: [-b ± √(b2 - 4*a*c)]/(2*a)
computer scince
ax2 + bx + c
ax2 + bx + c = y
Write the quadratic equation in the form ax2 + bx + c = 0 The roots are equal if and only if b2 - 4ac = 0. The expression, b2-4ac is called the [quadratic] discriminant.