Find the Roots of a Quadratic Equation in C Program

Find the Roots of a Quadratic Equation in C Program

How to write a c program to find the roots of a quadratic equation in c programming ?


C PROGRAMMING: SOLVING THE QUADRATIC EQUATION

Solution:
/* This program solves the quadratic equation */
#include <stdio.h>
#include <math.h>
int main()
{
double a, b, c;
double D;
double r1, r2;

printf("Enter the value of a: ");
scanf("%lf", &a);
printf("Enter the value of b: ");
scanf("%lf", &b);
printf("Enter the value of c: ");
scanf("%lf", &c);

D = b*b - 4*a*c;

if(D < 0)
printf("Roots are imaginary\n");
else
{
r1 = (-b + sqrt(D)) / (2*a);
r2 = (-b - sqrt(D)) / (2*a);

printf("Root 1 = %lf\n", r1);
printf("Root 2 = %lf\n", r2);
}

return 0;
}

[1] When you compile with gcc you need to link to the math library
gcc -o quadratic.e quadratic.c -lm
Otherwise, the compiler does not know about the square root function sqrt . You will also
need to use the -lm ‘compiler switch’ if you use other special functions: exp, sin, log, abs
[2] You should always check your codes as much as possible. Often all you can do is special
cases because you cannot solve the whole problem. In this case we don’t really need a
computer. Try, for example, a = 2,b = 5,c = −3 and then solve the problem with pencil
and paper (either with the quadratic equation or by factoring).
2
[3] Can anyone see a flaw in the program? What happens if you run the code with a =
2,b = 5,c = 4? What does ’nan’ mean? (Answer: not a number).
[4] What is the geometric interpretation of the discriminant being negative? Answer: the
parabola y = ax2 + bx + c doesn’t intersect the x-axis.
We can write a better program to deal with the case when the discriminant is negative.
/* This program solves the quadratic equation more completely*/
#include <stdio.h>
#include <math.h>
int main(void)
{
double a,b,c,root1,root2;
printf(" Please enter a \n");
scanf("%lf",&a);
printf(" Please enter b \n");
scanf("%lf",&b);
printf(" Please enter c \n");
scanf("%lf",&c);
if (b*b-4.*a*c>0)
{
root1 = (-b + sqrt(b*b-4.*a*c) ) / (2.*a);
root2 = (-b - sqrt(b*b-4.*a*c) ) / (2.*a);
printf("\n First root is %lf ",root1);
printf("\n Second root is %lf ",root2);
}
else
{
printf("\n Discriminant is negative! No roots!");
}
printf("\n ");
return 0;
}
3
[5] This upgraded program introduces to a new element of the C language: the if statement.
The syntax of the if statement is fairly straightforward: within the parentheses ( ) is an
arithmetic expression. If the expression is true, the lines within the first curly brackets { }
are executed. If it is false, the lines within the second curly brackets (following the else. )
are performed. More complex if statements are also available.
[6] Extra task for those interested: Could you write a program which can deal with negative
discriminants and print out a complex number answer like 2 + 3i? Hint: you do not need to
know how to use complex numbers in C to write the program.


Learn More :

Learn More Multiple Choice Question :