四次方程的根之和和根之積的絕對差?
本節我們將瞭解如何獲取四次方程中根之和與根之積的絕對差?
四次方程類似於 𝑎𝑥4+𝑏𝑥3+𝑐𝑥2+𝑑𝑥+𝑒
我們可以求解方程,然後嘗試透過一些正常步驟得到根之積和和,但這需要花費大量時間,而且這種方法效率不高。對於此類方程,我們有兩個公式。根之和始終為 −𝑏∕𝑎,根之積始終為 𝑒∕𝑎。因此我們必須僅找到 ∣−𝑏∕𝑎− 𝑒∕𝑎∣ ∣ 的值
演算法
rootSumProdDiff(a, b, c, d, e)
begin sum := -b/a prod := e/a return |sum - prod| end
示例
#include<iostream> #include<cmath> using namespace std; double rootSumProdDiff(double a, double b, double c, double d, double e){ double sum = double(-b/a); double prod = double(e/a); return abs(sum - prod); } main() { double a,b,c,d,e; cout << "Enter a, b, c, d, e for equation ax^4 + bx^3 + cx^2 + dx + e:"; cin >> a >> b >> c >> d >> e; cout << "Difference between sum and product of roots are: " << rootSumProdDiff(a, b, c, d, e); }
輸出
Enter a, b, c, d, e for equation ax^4 + bx^3 + cx^2 + dx + e:8 4 6 4 1 Difference between sum and product of roots are: 0.625
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