Determine the Number of Roots of Equations Involving nth Powers

Understanding the number of roots that a system of equations involving nth powers can have is crucial for solving complex mathematical problems. In this article, we delve into the nitty-gritty of solving a system defined by x^n y^n a and xy b. We will explore an in-depth analysis to determine the number of roots, considering both real and complex solutions.

Step-by-Step Analysis

Step 1: Analyze the Second Equation - The equation xy b can be rewritten as y b/x, assuming x ≠ 0. Step 2: Substitute into the First Equation - By substituting y b/x into the first equation x^n y^n a, we get: x^n (b/x)^n a x^n b^n / x^n a which simplifies to 1 a / b^n x^{2n} - ax^n b^n 0 Step 3: Letting z x^n - Let z x^n. The equation becomes: z^2 - az b^n 0 Step 4: Using the Quadratic Formula - This is a quadratic equation in the form z^2 - az b^n 0. Using the quadratic formula: z (a ± sqrt{a^2 - 4b^n}) / 2 Step 5: Analyzing the Discriminant - The discriminant D a^2 - 4b^n will determine the number of real roots: If D 0, there are 2 distinct real roots for z, leading to 2 values of x and hence 2 values of y'); If D 0, there is 1 real root for z, leading to 1 value of x and hence 1 value of y; If D 0, there are no real roots for z, meaning there are no real solutions for x and y. Step 6: Summary of the Number of Roots - Therefore, the number of roots of the original system of equations depends on the value of the discriminant: 2 real roots if a^2 - 4b^n 0 1 real root if a^2 - 4b^n 0 0 real roots if a^2 - 4b^n 0

Complex Solutions Considered

While real solutions are important, the possibility of complex solutions is also vital. Let's assume neither a nor b is zero. Then there are 2n roots or solutions (x', y'). Some may be degenerate, but we will skip that analysis. From the second equation, y b/x. Plugging into the first equation, we get:

x^n b x^{-n} a which simplifies to x^{2n} - ax^n b^n 0. This is a polynomial of order 2n with 2n roots, including complex. The final answers or roots would be the pairs x' 1/x' for all 2n solutions.

If b 0, then either x or y must be 0. If a is also 0, there is 1 solution x y 0. Otherwise, if y 0, there would be only n roots with x nth root of a. The final answers would thus be the n pairs x' 0.

Conclusion

The number of roots of the system can be 0, 1, or 2, depending on the values of a and b. Understanding this provides insight into the nature of the solutions and the complexity of the problem.