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CIE A-Level Maths Study Notes

2.9.3 Conjugate Pairs in Polynomial Equations

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+bia + bi and abi a - bi .
  • Polynomial Equations: These are equations of the form axn+bxn1++k=0 ax^n + bx^{n-1} + \ldots + k = 0 , where a,b,,k a, b, \ldots, k are real coefficients.

Application in Cubic and Quartic Equations

Cubic Equations

Example: Consider the cubic equation x36x2+11x6=0 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.

Cubic Equation

Image courtesy of Cuemath

Example

Equation: x33x2+5x3=0 x^3 - 3x^2 + 5x - 3 = 0

Solution:

1. Trial and Error Method:

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

2. Polynomial Division:

  • With a known root x=1 x = 1 , divide the cubic equation by (x1) (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 x = 1
  • Complex Roots: x=12ix = 1 - \sqrt{2}i and x=1+2i x = 1 + \sqrt{2}i

Quartic Equations

  • Example: For a quartic equation x410x3+35x250x+24=0 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.
Quadratic Equation

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

Example

Equation: x42x33x2+4x2=0 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=129462 x = \frac{1}{2} - \frac{\sqrt{9 - 4\sqrt{6}}}{2}
  • x=12+9462 x = \frac{1}{2} + \frac{\sqrt{9 - 4\sqrt{6}}}{2}
  • x=129+462 x = \frac{1}{2} - \frac{\sqrt{9 + 4\sqrt{6}}}{2}
  • x=12+9+462 x = \frac{1}{2} + \frac{\sqrt{9 + 4\sqrt{6}}}{2}
Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
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Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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