How to take Quadratic Equation as input via Scanf?

Question

I'm doing a course in C,where i was asked to find the roots of a quadratic equation.First i tried by hard-coding and it worked. Next i gave inputs using scanf(like a,b,c) it worked. But i failed in a scenario where the whole quadratic expression is taken as input i.e (ax^2+bx+c) and retrieve these a,b,c values from the expression. I spent a lot of time on this,searched online i couldn't find the answer so i am asking here for help.

#include<stdio.h>
#include<stdlib.h>
#include<math.h>

#define PI 3.14

int main(void)
{

puts("---- ROOTS ----");

char equ[20]; //Quadratic Expression in an Array 
float a,b,c,ope;
float root1,root2;

printf("please provide the expression :");
scanf("%d",&equ[20]);//Example : 5x^2+3x+1 as input

a == equ[0];//since ax^2+bx+c in the above expression a==5
b == equ[3];//b==3
c == equ[6];//c==1

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

printf("The root 1 of the expression is : %d", root1); 
printf("\nThe root 2 of the expression is : %d", root2);

return EXIT_SUCCESS;
}

OUTPUT :

PS F:\Thousand C GO> gcc 3.c
PS F:\Thousand C GO> ./a
---- ROOTS ----
please provide the expression :5x^2+3x+1//edited
The root 1 of the expression is : 0
The root 2 of the expression is : 0  

I wanted to know if there is a way to solve this problem in C,if so how? if not why?.

Help is much appreciated. Thanks.

Answer 1

1. To read the string -

try

scanf("%s",equ);

or

scanf("%s",&equ[0]);

2. == is the 'equal to' operator. You may want to change this to = (assignment operator)

3. Recheck the index values that you are using to extract the coefficients in the quadratic equation.

4. Coefficients are stored in ascii values in the character array. You need to convert them to appropriate integer or numeric value.

Answer 2

Hi here is the changed code :

scanf("%s",equ);//Example : 5x^2+3x+1 as input NOT

a = equ[0]-48;//since ax^2+bx+c in the above expression a==5
b = equ[5]-48;//b==3
c = equ[8]-48;//c==1
//printf("\n%f %f %f",a,b,c);

ope = sqrt(b*b -4*a*c);
printf("\n%f",ope);
root1 = (-b + ope)/(2*a);
root2 = (-b - ope)/(2*a);

printf("\nThe root 1 of the expression is : %f", root1); 
printf("\nThe root 2 of the expression is : %f", root2);

Now I will try to address the problems

1. try to use the correct placeholder for inputting character array which is "%s" not "%d".
2. don't give scanf("%d",&equ[20]) for scanning character array give scanf("%s",equ) only. here is the explanation.
3. the index values are wrong correct them.
4. the characters are stored as ASCII values. here is a link of all the ASCII values note that the ASCII value of 1 is 49, so while converting to float (implicit typecasting), convert to the appropriate value. In your case subtract 48.
5. == is a comparison operator, change it to assignment operator =.
6. put a bracket over (-b + ope)/(2*a), because your /2 evaluates first that *a is evaluated, this is operator precedence and associativity .
7. now the major problem, Notice that I have changed the equation to 5x^2+3x+1, this is because in the case of 5x^2+3x+2 the value of b*b -4*a*c will be -31 which is negative and you can not store an imaginary number in a float type.here and here are some good reads for this topic.

It will take some time to grasp all this, but don't give up.

Happy Coding!

Edit: As mentioned in point 1 the right placeholder should be used for the corresponding datatype.

8. In the last two lines, I have changed the %d to %f for printing root1 and root2 which are float and NOT int. here is a good read for very beginner on the same.