Understanding of conjugate pairs in polynomial equations is essential. This topic explores the principle that non-real roots of polynomials with real coefficients occur in conjugate pairs, particularly focusing on cubic and quartic equations.

Introduction to Conjugate Pairs

• Complex Roots: In any polynomial with real coefficients, if a complex number is a root, its conjugate will also be a root.
• Conjugate Pair: A pair of complex numbers of the form $a + bi$ and $a - bi$.
• Polynomial Equations: These are equations of the form $ax^n + bx^{n-1} + \ldots + k = 0$, where $a, b, \ldots, k$ are real coefficients.

Application in Cubic and Quartic Equations

Cubic Equations

Example: Consider the cubic equation $x^3 - 6x^2 + 11x - 6 = 0$.

Roots: This equation has three real roots: 1, 2, and 3.

Conjugate Pairs: Since all roots are real, there are no conjugate pairs in this case.

Image courtesy of Cuemath

Example

Equation: $x^3 - 3x^2 + 5x - 3 = 0$

Solution:

1. Trial and Error Method:

• Test simple numbers $0, 1, -1, etc.$ to find roots.
• Here, $x = 1$ works (equation equals 0), so 1 is a root.

2. Polynomial Division:

• With a known root $x = 1$, divide the cubic equation by $(x - 1)$ to get a quadratic equation.

3. Solving the Quadratic Equation:

• Use the quadratic formula to find the other two roots.

Roots Found:

• Real Root: $x = 1$
• Complex Roots: $x = 1 - \sqrt{2}i$ and $x = 1 + \sqrt{2}i$

Quartic Equations

• Example: For a quartic equation $x^4 - 10x^3 + 35x^2 - 50x + 24 = 0$.
• Roots: The roots of this equation are 1, 2, 3, and 4.
• Conjugate Pairs: Similar to the cubic example, this equation also has only real roots, hence no conjugate pairs.

The graph provides a visual representation of the function in the range $x = -2$ to $x = 6$.

Example

Equation: $x^4 - 2x^3 - 3x^2 + 4x - 2 = 0$

Solution:

1. Trial and Error Method:

• Test simple values for roots. This equation has no obvious roots from this method.

2. Algebraic Techniques:

• For quartic equations, consider factoring, completing the square, or specific quartic formulas.
• These methods can be intricate, involving multiple steps.

3. Finding Real Roots:

• The equation simplifies to a form where roots involve square roots.

Roots Found:

• $x = \frac{1}{2} - \frac{\sqrt{9 - 4\sqrt{6}}}{2}$
• $x = \frac{1}{2} + \frac{\sqrt{9 - 4\sqrt{6}}}{2}$
• $x = \frac{1}{2} - \frac{\sqrt{9 + 4\sqrt{6}}}{2}$
• $x = \frac{1}{2} + \frac{\sqrt{9 + 4\sqrt{6}}}{2}$