Unit 5: Theory of Equations: A Comprehensive Study Resource

The theory of equations is a fundamental branch of algebra that deals with polynomial equations, their roots, and various properties. It explores methods to find roots, understand their nature, and relate them to the coefficients of the polynomial.

Fundamental Theorem of Algebra

Statement of the Fundamental Theorem of Algebra

Every non-constant single-variable polynomial with complex coefficients has at least one complex root. This includes polynomials with real coefficients, as real numbers are a subset of complex numbers.

Consequences of the Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra has several important consequences that are crucial for understanding polynomial equations:

1. Existence of Roots: It guarantees that every polynomial equation of degree n (n ≥ 1) has at least one root in the complex number system.
2. Number of Roots: A polynomial equation of degree n has exactly n roots in the complex number system, provided each root is counted according to its multiplicity. For example, x² - 2x + 1 = 0 has roots x=1 with multiplicity 2.
3. Factorization into Linear Factors: If P(x) is a polynomial of degree n, then it can be factored into n linear factors over the complex numbers: P(x) = a(x - α₁)(x - α₂)...(x - α_n), where α₁, α₂, ..., α_n are the roots and 'a' is the leading coefficient.
4. Conjugate Pairs Theorem: For a polynomial equation with real coefficients, if a complex number (a + bi) is a root, then its conjugate (a - bi) must also be a root. This means complex roots always occur in conjugate pairs.

Relation Between Roots and Coefficients

Vieta's formulas establish a direct relationship between the roots of a polynomial equation and its coefficients. These relations are incredibly useful for finding unknown roots or for constructing polynomial equations.

General Polynomial Equation

Consider a general polynomial equation of degree n:

a_nxⁿ + a_n-1x^n-1 + ... + a₁x + a₀ = 0, where a_n ≠ 0.

If α₁, α₂, ..., α_n are the n roots of this equation, then:

• Sum of roots: ∑α_i = α₁ + α₂ + ... + α_n = -a_n-1 / a_n
• Sum of products of roots taken two at a time: ∑α_iα_j = (a_n-2) / a_n
• Sum of products of roots taken three at a time: ∑α_iα_jα_k = -(a_n-3) / a_n
• ...
• Product of all roots: α₁α₂...α_n = (-1)ⁿ(a₀) / a_n

For Quadratic Equations

Let the quadratic equation be ax² + bx + c = 0, and its roots be α and β.

• Sum of roots: α + β = -b / a
• Product of roots: αβ = c / a

For Cubic Equations

Let the cubic equation be ax³ + bx² + cx + d = 0, and its roots be α, β, and γ.

• Sum of roots: α + β + γ = -b / a
• Sum of products of roots taken two at a time: αβ + βγ + γα = c / a
• Product of roots: αβγ = -d / a

For Biquadratic Equations

Let the biquadratic equation be ax⁴ + bx³ + cx² + dx + e = 0, and its roots be α, β, γ, and δ.

• Sum of roots: α + β + γ + δ = -b / a
• Sum of products of roots taken two at a time: αβ + αγ + αδ + βγ + βδ + γδ = c / a
• Sum of products of roots taken three at a time: αβγ + αβδ + αγδ + βγδ = -d / a
• Product of roots: αβγδ = e / a

Observation: These relations are extremely useful when some information about the roots is known, allowing us to find other roots or conditions on coefficients without directly solving the equation.

Solutions of Cubic and Biquadratic Equations When Some Conditions on Roots of Equations are Given

When specific conditions or relations among the roots are provided, solving cubic and biquadratic equations becomes much simpler. These conditions allow us to reduce the number of unknowns or simplify the use of Vieta's formulas.

Cubic Equations (ax³ + bx² + cx + d = 0)

Common conditions on roots (α, β, γ) and approaches:

• One root is known (e.g., α = k):
  1. Since k is a root, P(k) = 0. This can be used to find an unknown coefficient if present.
  2. Divide the polynomial P(x) by (x - k) using synthetic division or long division. The resulting quotient will be a quadratic equation.
  3. Solve the quadratic equation to find the remaining two roots (β and γ).
• Roots are in Arithmetic Progression (A.P.):
  1. Assume the roots are (a - d), a, (a + d).
  2. Use the sum of roots relation: (a - d) + a + (a + d) = -b/a => 3a = -b/a. This directly gives the middle root 'a'.
  3. Substitute 'a' back into the equation, or use the product of roots relation and the sum of products of roots taken two at a time to find 'd' and then the other roots.
• Roots are in Geometric Progression (G.P.):
  1. Assume the roots are (a/r), a, (ar).
  2. Use the product of roots relation: (a/r) * a * (ar) = -d/a => a³ = -d/a. This directly gives the middle root 'a'.
  3. Substitute 'a' back into the equation, or use the sum of roots relation and the sum of products of roots taken two at a time to find 'r' and then the other roots.
• Two roots are equal (e.g., α = β):
  1. Assume the roots are α, α, γ.
  2. Use Vieta's formulas: 2α + γ = -b/a; α² + 2αγ = c/a; α²γ = -d/a.
  3. Solve this system of equations to find α and γ. Alternatively, if a root is repeated, then it is also a root of the derivative of the polynomial P'(x). Find common roots of P(x) and P'(x).
• Sum of two roots is zero (e.g., α + β = 0 => β = -α):
  1. Assume the roots are α, -α, γ.
  2. Use the sum of roots relation: α + (-α) + γ = -b/a => γ = -b/a. This directly gives one root.
  3. Divide the polynomial by (x - γ), and solve the resulting quadratic (x² - α² = 0) to find α and -α.

Biquadratic Equations (ax⁴ + bx³ + cx² + dx + e = 0)

Common conditions on roots (α, β, γ, δ) and approaches:

• One or two roots are known:
  1. If one root 'k' is known, divide P(x) by (x - k) to get a cubic equation.
  2. If two roots 'k₁', 'k₂' are known, divide P(x) by (x - k₁)(x - k₂) to get a quadratic equation.
  3. Solve the resulting lower-degree polynomial.
• Roots are in Arithmetic Progression (A.P.):
  1. If the roots are symmetric, assume them as (a - 3d), (a - d), (a + d), (a + 3d).
  2. Use the sum of roots relation: (a - 3d) + (a - d) + (a + d) + (a + 3d) = -b/a => 4a = -b/a. This gives 'a'.
  3. Substitute 'a' and use other Vieta's relations to find 'd'.
  4. If not symmetric, assume as a, a+d, a+2d, a+3d. This makes calculations more complex but still solvable with Vieta's formulas.
• Roots are in Geometric Progression (G.P.):
  1. If the roots are symmetric, assume them as (a/r³), (a/r), (ar), (ar³).
  2. Use the product of roots relation: (a/r³)(a/r)(ar)(ar³) = e/a => a⁴ = e/a. This gives 'a'.
  3. Substitute 'a' and use other Vieta's relations to find 'r'.
  4. If not symmetric, assume as a, ar, ar², ar³.
• Pairs of roots are equal:
  1. e.g., roots are α, α, β, β.
  2. Use Vieta's formulas: 2α + 2β = -b/a; α² + β² + 4αβ = c/a; 2α²β + 2αβ² = -d/a; α²β² = e/a.
  3. Solve this system of equations.
• Sum of two roots is zero (e.g., α + β = 0 => β = -α):
  1. Assume the roots are α, -α, γ, δ.
  2. The sum of roots becomes γ + δ = -b/a.
  3. The product of roots becomes -α²γδ = e/a.
  4. If the polynomial is P(x), then P(x) can be divided by (x - α)(x + α) = (x² - α²), leaving a quadratic for γ, δ.
  5. Use the fact that a polynomial with roots α, -α, γ, δ can be written as (x² - α²)(x² - (γ+δ)x + γδ) = 0. Equate coefficients.
• Product of two roots is equal to the product of the other two (e.g., αβ = γδ):
  1. Let αβ = k and γδ = k.
  2. From Vieta's formulas, αβγδ = e/a => k² = e/a. This gives k.
  3. The equation can often be expressed as (x² + px + k)(x² + qx + k) = 0 by comparing coefficients.
• Reciprocal Equations:
  1. A reciprocal equation is one where if 'k' is a root, then '1/k' is also a root. The coefficients are symmetric (e.g., ax⁴ + bx³ + cx² + bx + a = 0).
  2. If the degree is even (e.g., 4), divide the entire equation by x².
  3. Rearrange terms to group (x² + 1/x²) and (x + 1/x).
  4. Substitute y = x + 1/x, then y² = x² + 2 + 1/x² => x² + 1/x² = y² - 2.
  5. This transforms the biquadratic equation into a quadratic equation in 'y', which can be solved. Once 'y' is found, solve for 'x' using x + 1/x = y.