Absolute difference between sum and product of roots of a quartic equation?

In this section we will see how to find the absolute difference between the sum of the roots and the product of the roots of a quartic equation. A quartic equation has the form ax⁴ + bx³ + cx² + dx + e = 0.

We can solve the equation and then calculate the sum and product of the roots using traditional methods, but that approach is time-consuming and inefficient. For a quartic equation, we have two important formulas −

• Sum of roots = -b/a
• Product of roots = e/a

Therefore, we need to find the value of |(-b/a) - (e/a)|.

Syntax

double rootSumProdDiff(double a, double b, double c, double d, double e);

Algorithm

begin
   sum := -b/a
   prod := e/a
   return |sum - prod|
end

Example

Here's a complete C program to calculate the absolute difference between sum and product of roots −

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

double rootSumProdDiff(double a, double b, double c, double d, double e) {
    double sum = -b / a;
    double prod = e / a;
    return fabs(sum - prod);
}

int main() {
    double a = 8, b = 4, c = 6, d = 4, e = 1;
    
    printf("Quartic equation: %.0fx^4 + %.0fx^3 + %.0fx^2 + %.0fx + %.0f = 0<br>", 
           a, b, c, d, e);
    
    double sum = -b / a;
    double prod = e / a;
    
    printf("Sum of roots: %.3f<br>", sum);
    printf("Product of roots: %.3f<br>", prod);
    printf("Absolute difference: %.3f<br>", rootSumProdDiff(a, b, c, d, e));
    
    return 0;
}

Quartic equation: 8x^4 + 4x^3 + 6x^2 + 4x + 1 = 0
Sum of roots: -0.500
Product of roots: 0.125
Absolute difference: 0.625

Key Points

• For any quartic equation ax⁴ + bx³ + cx² + dx + e = 0, the sum of roots is always -b/a
• The product of roots is always e/a
• We use fabs() function to get the absolute difference
• This method avoids the complex process of finding individual roots

Conclusion

Using Vieta's formulas for quartic equations provides an efficient way to calculate the absolute difference between sum and product of roots without solving for individual roots. This approach has O(1) time complexity.