Write the C program that solve the quadratic equation problem

The output should appear as follows:

Enter A: 3
Enter B: 5
Enter C: 2
There are two solutions: -0.666667 and -1.
Press any key to continue_

thank you,

Of course we're not gonna do this, unless you pay me \$20.00 dollars for this problem.

But to start you off, this is the quadratic equation :

``x = -b/(2a) +-  sqrt(b^2 - 4ac) / (2a).``

Since they gave you a,b, and c. You just have …

Your problem is assuming that x1 and x2 are integers when you print them out.
Additionally, a,b,c are all likely to be floating point numbers. So read them in as floats. So make your output statement `printf("%g and %g\n",x1,x2);` Also you can use the sqrt function in line `k=sqrt(d);` .

## All 7 Replies

Of course we're not gonna do this, unless you pay me \$20.00 dollars for this problem.

But to start you off, this is the quadratic equation :

``x = -b/(2a) +-  sqrt(b^2 - 4ac) / (2a).``

Since they gave you a,b, and c. You just have to plug it in though.

A word of caution this part of the equation b^2 - 4ac is called the discriminant. Make sure this is positive else the solution is imaginary.

Below is my code for the problem, please correct my mistake:

``````#include <stdafx.h>
#include "simpio.h"
#include "genlib.h"
#include "math.h"
main()
{
float a,b,c,d,k,x1,x2;
float pow(float,float);

printf("Enter A: ");
a=GetInteger();
printf("Enter B: ");
b=GetInteger();
printf("Enter C: ");
c=GetInteger();
d=(b*b-4*a*c);
k=pow(d,1./2);
x1=(-b+k)/(2*a);
x2=(-b-k)/(2*a);
printf("%d and %d",x1,x2);
}``````

Your problem is assuming that x1 and x2 are integers when you print them out.
Additionally, a,b,c are all likely to be floating point numbers. So read them in as floats. So make your output statement `printf("%g and %g\n",x1,x2);` Also you can use the sqrt function in line `k=sqrt(d);` .

HOWEVER: Can I just point out that using the formula given is a horrible way to calculate the roots to a quadratic equation due to the numeric instability of the solution. The "better" way is to calculate x1,x2 and determine which biggest root
(either in terms of magnitude or accuracy as a solution) and use the relationship `x1*x2=c/a` to get the other root.

There are further tricks to getting better roots from a quadratic equation, particularly if the roots are close, if b^2-4ac is near zero ,e.g a Newton Raphson step etc.

#include<stdio.h>
#include<iostream.h>
#include<math.h>
main()
{
float a,b,c,d,k,x,y;
printf("Enter the values of a,b,c:");
scanf("%f%f%f",&a,&b,&c);
d = ((b*b)-(4*a*c));
if(d>=0)
{
k = sqrt(d);
x = (-b+k)/(2*a);
y = (-b-k)/(2*a);
printf("The quadratic equation is x = %f and y = %f",x,y);
}
else
{
float re, im;
d*=-1;
k = sqrt(d);
re = -b/(2*a);
im = k/(2*a);
printf("The quadratic equation equation is y=%f-(%fi)\n",re,im);
}
return 0;
}

commented: No code tags; resurrecting old thread; posting an alledged answer in full to help him cheat. Good Job :-( -4

@moseshenry
That's not what this forum for... You should read the forum rules before you post the code like that...

Write a C porgramming for the first line of the data set for this problem is an integer that represents the number of data sets that follow and Each data set is on a separate line and consists of 2 times in the form hh:mm:ss.Values less then 10 have a leading zero.

hello all
can help me to writing c code for this eq. please

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