Solving Polynomial Equations: Descartes and Cardan Methods
1. Nature of Roots: Descartes’s Rule of Signs
Find the nature of the roots for the equation: x⁴ + 2x² + 3x – 2 = 0
Applying Descartes’ Rule of Signs
Given the function: f(x) = x⁴ + 2x² + 3x – 2
Signs of coefficients: +, +, +, –
Count sign changes:
- + to + (No change)
- + to + (No change)
- + to – (1 change)
Therefore, there is exactly one positive real root.
Replace x with -x:
f(-x) = (-x)⁴ + 2(-x)² + 3(-x) – 2 = x⁴ + 2x² – 3x – 2
Signs: +, +, -, –
Count sign changes:
- + to + (No change)
- + to – (1 change)
- – to – (No change)
Therefore, there is exactly one negative real root.
Summary of roots:
- Degree of equation = 4 (Total roots)
- 1 positive real root
- 1 negative real root
- Remaining roots: 4 – 2 = 2 (Must be complex conjugate roots)
Conclusion: The equation has 1 positive, 1 negative, and 2 complex roots.
2. Solving Cubic Equations via Cardan’s Method
Solve: x³ + 9x² + 15x – 25 = 0
Step 1: Convert to Depressed Cubic
To remove the x² term, substitute x = y – (9/3) = y – 3.
Step 2: Substitution and Expansion
Substitute x = y – 3 into the equation:
- (y – 3)³ = y³ – 9y² + 27y – 27
- 9(y – 3)² = 9y² – 54y + 81
- 15(y – 3) = 15y – 45
Step 3: Combine Terms
Simplifying the expression results in the depressed cubic:
y³ – 12y – 16 = 0
Step 4: Standard Form
Using y³ + py + q = 0, we identify p = -12 and q = -16.
Step 5: Apply Cardan’s Formula
Calculate the discriminant Δ:
Δ = (q/2)² + (p/3)³ = (-8)² + (-4)³ = 64 – 64 = 0
Step 6: Solve for y
Since Δ = 0, we have real repeated roots:
y = ∛(-q/2 + √Δ) + ∛(-q/2 – √Δ) = ∛8 + ∛8 = 2 + 2 = 4
Step 7: Back Substitution
x = y – 3 = 4 – 3 = 1
Step 8: Find Other Roots
Dividing the polynomial by (x – 1) yields (x² + 10x + 25), which is (x + 5)².
Final Answer
The roots are: x = 1, x = -5, x = -5
- One simple root: 1
- One repeated root: -5