# C Program to Find Roots of a Quadratic Equation

C program to find root of a quadratic equation; In this tutorial, i am going to show you how to find the root of a quadratic equation in c programs.

Here, i will show you the standard form of a quadratic equation is:

```ax2 + bx + c = 0, where
a, b and c are real numbers and
a != 0
```

The term `b2; - 4ac` is known as the discriminant of a quadratic equation. It tells the nature of the roots.

• If the discriminant is greater than `0`, the roots are real and different.
• If the discriminant is equal to `0`, the roots are real and equal.
• If the discriminant is less than `0`, the roots are complex and different.

## Algorithm and Programs To Find Roots of a Quadratic Equation

• Algorithm to find roots of a quadratic equation
• C Program to Find Roots of a Quadratic Equation using if else
• C Program to Find Roots of a Quadratic Equation using switch case

### Algorithm to find roots of a quadratic equation

Follow the bellow given algorithm steps to write a c program to find the roots of a quadratic equation ax2 + bx + c = 0; as follows:

1. Start Program
2. Read the value of a, b, c.
3. Calculate k = b*b – 4*a*c.
4. If (d < 0) Display “Roots are Imaginary, calculater1 = (-b +i ? k)/ 2a and r2 =(b + i? k)/ 2a. else if (d = 0) Display “Roots are Equal” and calculate r1 = r2 = (-b / 2*a) …
5. Print r1 and r2.
6. End Program

### C Program to Find Roots of a Quadratic Equation using if else

```/**
* C program to find all roots of a quadratic equation
*/
#include <stdio.h>
#include <math.h> /* Used for sqrt() */
int main()
{
float a, b, c;
float root1, root2, imaginary;
float discriminant;

printf("Enter values of a, b, c of quadratic equation (aX^2 + bX + c): ");
scanf("%f%f%f", &a, &b, &c);

/* Find discriminant of the equation */
discriminant = (b * b) - (4 * a * c);

/* Find the nature of discriminant */
if(discriminant > 0)
{
root1 = (-b + sqrt(discriminant)) / (2*a);
root2 = (-b - sqrt(discriminant)) / (2*a);
printf("Two distinct and real roots exists: %.2f and %.2f", root1, root2);
}
else if(discriminant == 0)
{
root1 = root2 = -b / (2 * a);
printf("Two equal and real roots exists: %.2f and %.2f", root1, root2);
}
else if(discriminant < 0)
{
root1 = root2 = -b / (2 * a);
imaginary = sqrt(-discriminant) / (2 * a);
printf("Two distinct complex roots exists: %.2f + i%.2f and %.2f - i%.2f",
root1, imaginary, root2, imaginary);
}
return 0;
}```

The result of the above c program; as follows:

```Enter values of a, b, c of quadratic equation (aX^2 + bX + c): 8 -4 -2
Two distinct and real roots exists: 0.81 and -0.31```

### C Program to Find Roots of a Quadratic Equation using switch case

```#include <stdio.h>
#include<math.h>
int main()
{
float a, b, c;
float root1, root2, imaginary, discriminant;

printf("\n Please Enter values of a, b, c of Quadratic Equation :  ");
scanf("%f%f%f", &a, &b, &c);

discriminant = (b * b) - (4 * a *c);

switch(discriminant > 0)
{
case 1:
root1 = (-b + sqrt(discriminant) / (2 * a));
root2 = (-b - sqrt(discriminant) / (2 * a));
printf("\n Two Distinct Real Roots Exists: root1 = %.2f and root2 = %.2f", root1, root2);
break;
case 0:
switch(discriminant < 0)
{
case 1: //True
root1 = root2 = -b / (2 * a);
imaginary = sqrt(-discriminant) / (2 * a);
printf("\n Two Distinct Complex Roots Exists: root1 = %.2f+%.2f and root2 = %.2f-%.2f", root1, imaginary, root2, imaginary);
break;
case 0: // False (Discriminant = 0)
root1 = root2 = -b / (2 * a);
printf("\n Two Equal and Real Roots Exists: root1 = %.2f and root2 = %.2f", root1, root2);
break;
}
}

return 0;
}```

The result of the above c program; as follows:

```Please Enter values of a, b, c of Quadratic Equation :  2 3 5
Two Distinct Complex Roots Exists: root1 = -0.75+1.39 and root2 = -0.75-1.39```

Categories C