In this blog post, we learn how to write a C program to find all roots of a quadratic equation using switch case?. We will write the C program to find all roots of a quadratic equation using switch case. Write C program to find roots of quadratic equations using switch statements. How to find all roots of a quadratic equation using if else in C programming. Logic to find roots of quadratic equation in C programming.
Let see an example,
Input : a = 1, b = -2, c = 1
Output: Roots are real and same
1
Input : a = 1, b = 7, c = 12
Output: Roots are real and different
-3, -4
Input : a = 1, b = 1, c = 1
Output : Roots are complex
-0.5 + i1.73205
-0.5 - i1.73205
Quadratic equation:
In algebra, a quadratic equation is an equation that can be rearranged in standard form as,
ax2 + bx + c = 0
Below is a direct formula for finding the roots of the quadratic equation.
There are the following important cases of this formula.
Case 1: (Discriminant < 0)
b*b < 4*a*c, then roots are complex(not real). For example roots of x2 + x + 1, roots are -0.5 + i1.73205 and -0.5 - i1.73205
Case 2: (Discriminant == 0)
b*b == 4*a*c, then roots are real and both roots are the same. For example, roots of x2 - 2x + 1 are 1 and 1
Case 3: (Discriminant > 0)
b*b > 4*a*c, then roots are real and different. For example, roots of x2 - 7x - 12 are 3 and 4
Logic to find roots of quadratic equation using switch case:
- Ask the user to enter the value of a, b, and C.
- Calculate the discriminant value using the a, b, and c.
- Compute the roots based on the nature of the discriminant.
- The expression (discriminant > 0) can have two possible cases i.e. case 0 and case 1.
- For case 1 ( when discriminant +ve) compute the roots as per the above-mentioned formula.
- For case 0, again there are two scenario discriminant is -ve (discriminant < 0)or discriminant is zero (discriminant == 0).
C program to find the roots of a quadratic equation using switch case:
The below program ask the user to enter the value of a,b and c. After getting the value from the user it will calculate on the basis of ‘Discriminant’ value using switch case.
#include <stdio.h>
#include <math.h>
int main()
{
float a, b, c;
float root1, root2, imaginary, discriminant;
printf("Enter value of 'a' of quadratic equation (aX^2 + bX + c): ");
scanf("%f", &a);
printf("Enter value of 'b' of quadratic equation (aX^2 + bX + c): ");
scanf("%f",&b);
printf("Enter values of 'c' of quadratic equation (aX^2 + bX + c): ");
scanf("%f",&c);
// Find discriminant of the equation
discriminant = (b * b) - (4 * a * c);
switch(discriminant > 0)
{
case 1:
// If discriminant is positive
root1 = (-b + sqrt(discriminant)) / (2 * a);
root2 = (-b - sqrt(discriminant)) / (2 * a);
printf("Two distinct and real roots exists: %.2f and %.2f",
root1, root2);
break;
case 0:
// If discriminant is not positive
switch(discriminant < 0)
{
case 1:
// If discriminant is negative
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);
break;
case 0:
// If discriminant is zero
root1 = root2 = -b / (2 * a);
printf("Two equal and real roots exists: %.2f and %.2f", root1, root2);
break;
}
}
return 0;
}
Output:
Enter value of a of quadratic equation (aX^2 + bX + c): 2
Enter value of b of quadratic equation (aX^2 + bX + c): 7
Enter values of c of quadratic equation (aX^2 + bX + c): 2
Two distinct and real roots exist: -0.31 and -3.19
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