Bartlomiej Mlynarkiewicz - 1 year ago 93
C Question

# How to solve quadratic equations in C, using 3 coefficients?

My professor told us to write a c program which solves quadratic equations but he added that at the start he wants us to define 4 sets of 3 coefficients for each a, b and c. So in other words I have to define 4 sets of 3 different coefficients for each a, b, and c and once the program solves the equation for the 1st set of 3 coefficients for a,b and c its to proceed and solve the next set of 3 defined coefficients until all 4 sets are solved.

I am able to write a program that solves the quadratic equation by defining each a,b and c coefficient individual using scanf. Can you please help, as I can't find the answer anywhere?

Thats what I have written so far, its pretty simple but it works. I'm in an engineering course and we only have 2hrs a week of labs which dont reflect the lectures.

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

double a,b,c, root1, root2, discriminant, realpart, imaginarypart;
char anwser;
char x [100] = "Are your choices correct ? Y /N ";
int main ()
{
start:
printf("Enter your coefficient for a : ");
scanf("%lf",&a);
printf("Enter your coefficient for b : ");
scanf("%lf",&b);
printf("Enter your coefficient for c : ");
scanf("%lf",&c);
printf("Your coefficients are : \na = %lf\nb = %lf\nc = %lf\n",a,b,c);
printf("%s",x );
scanf("%s",&anwser);
if (!strcmp(&anwser, "N"))
{
goto start;
}

discriminant = (b*b - 4*a*c);
root1 = (-b - sqrt(discriminant)) / (2*a);
root2 = (-b + sqrt(discriminant)) / (2*a);

if (discriminant > 0 )
{
printf("root1 = %lf\nroot2 = %lf\n", root1, root2);
}
else if (discriminant == 0)
{
printf("root1 = root2 = %lf\n",root1);
}
else
{
realpart = -b/(2*a);
imaginarypart = sqrt(-discriminant)/(2*a);
printf("root1 = %lf + %lfi\nroot2 = %lf - %lfi\n",realpart, imaginarypart, realpart, imaginarypart);
}
}
``````

You mix 2 things: defining 4 sets and reading them from input (as I understood from your "using `scanf`").

(Reading them from input makes no problem as new read values overwrite old.)

Defining 4 sets in advance may be simply done by arrays:

``````float as[4], bs[4], cs[4];        // as[0], bs[0], cs[0] is the 1st set, etc.
float a    , b    , c    ;

// Assigning values to sets: as[0] = ...; bs[0] = ..., cs[0] = ...;
as[1] = ...; bs[1] = ..., cs[1] = ...;
......................................
......................................

//  or initialize them directly in the previous declaration, e.g.
//                           float as[] = {1,  4, -2,  3},
//                                 bs[] = {0,  5,  1, -1},
//                                 cs[] = {2, -4,  1,  2};

for (int i; i < 4; ++i)
{
a = a[i];
b = b[i];
c = c[i];
// Code (or a function call) for computing and printing result(s) from a, b, c
}
``````
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