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#### Roots of Equations

###### Coefficients of Equations ###### Algebraic Combinations
• Finding an equation through algebraic manipulation
###### Examples
1. x³ + 3x² – 2x + 5 = 0 has roots α, β, γ. Find an equation with roots (α – 1), (β – 1), (γ – 1)
• Using coefficients:
• α + β + γ = -3
• αβ + βγ + αγ = -2
• αβγ = -5
• Equation to find:
• (α – 1) + (β – 1) + (γ – 1) = α + β + γ – 3
• ∴ -3 – 3 = 6
• (α – 1)(β – 1) + (α – 1)(γ – 1) + (β – 1)(γ – 1)
• = αβ – α – β + 1 + αγ – α – γ + 1 + βγ – β – γ + 1
• = (αβ + αγ + βγ) – 2(α + β + γ) + 3
• ∴ -2 – 2(-3) + 3 = 7
• (α – 1)(β – 1)(γ – 1)
• = (αβ – α – β + 1)(γ – 1)
• = αβγ – αβ – αγ + α – βγ + β + γ – 1
• = αβγ – (αβ + αγ + βγ) + (α + β + γ) – 1
• ∴ -5 – (-2) + (-3) + 1 = -7
• Thus the equation is x³ + 6x² + 7x + 7 = 0
###### Substitution
• For finding equations whose roots have been raised by a power or undergone common algebraic manipulation
###### Examples
1. 8x³ + 12x² + 4x – 1 = 0 has roots α, β, γ. Find the equation with roots (2α + 1), (2β + 1), (2γ + 1)
• Let y = 2x + 1, therefore, x = y – 1/2
• Substitute into the equation:
• 8(y – 1/2)³ + 12(y – 1/2)² + 4(y-1/2) – 1 = 0
• 8(y – 1)³/8 + 12(y – 1)²/4 + 4(y – 1)/2 – 1 = 0
• (y – 1)³(y² – 2y + 1) + 3(y² – 2y + 1) + 2y – 2 – 1 = 0
• y³ – 2y² + y – y² + 2y – 1 + 3y² – 6y + 3 + 2y – 2 – 1 = 0
• y³ – y – 1 = 0
• Thus the equation is x³ – x – 1 = 0
###### Recurrence Relations
• For finding sums of roots to a specific degree of power
###### Examples
1. For the equation found above (x³ – x – 1 = 0), find S-2 where Sn = (2α + 1)n + (2β + 1)n + (2γ + 1)n
• x³ – x – 1 = 0 has roots (2α + 1), (2β + 1) and (2γ + 1)
• Substitute each root into the equation, e.g.
• (2α + 1)3 + (2α + 1) – 1 = 0
• Add all the equations formed together:
• (2α + 1)3 + (2β + 1)3 + …
• We get S3 – S1 – 3 = 0
• We must fins S-2, so if we multiply with S-1 it would result in:
• S2 – 3 – S-1 = 0
• S-1 = S2 – 3
• By repeating the process (multiply with S-1), we get:
• S-2 = S1 – S-1
• By manipulating equations, we get a useful equation:
• S-2 = S1 – S2 + 3
• We know S1 = 1 but to find S2, use substitution
• Let y = x²
• ∴ x = √y
• By substituting into the initial equation
• (√y)³ – √y – 1 = 0
• √y(y) – √y – 1 = 0
• √y(y – 1) = 1
• y(y – 1)² = 1
• y³ – 2y² + y – 1 = 0
• Thus S2 = 2
• Thus S-2 = 1 – 2 + 3 = 2