A-LevelFurther MathematicsRoots of polynomial equations9231/12 · May/June 2013 · Q38 Marks
The cubic equation x³ – 2x² – 3x + 4 = 0 has roots α, β, γ. Given that c = α + β + γ, state the value of c. Use the substitution y = c − x to find a cubic equation whose roots are α + β, β + γ, γ + α. Find a cubic equation whose roots are 1/(α + β), 1/(β + γ), 1/(γ + α). Hence evaluate 1/(α + β)² + 1/(β + γ)² + 1/(γ + α)².
- (a)The cubic equation x³ – 2x² – 3x + 4 = 0 has roots α, β, γ. Given that c = α + β + γ, state the value of c.[1]
- (b)Use the substitution y = c − x to find a cubic equation whose roots are α + β, β + γ, γ + α.[3]
- (c)Find a cubic equation whose roots are 1/(α + β), 1/(β + γ), 1/(γ + α).[2]
- (d)Hence evaluate 1/(α + β)² + 1/(β + γ)² + 1/(γ + α)².[2]
Mark scheme
Marking points from the official mark scheme (May/June 2013). The mark or mark type is shown in brackets.
- B1 Uses Σα = -b/a. c = 2
- M1 Uses substitution (α + β = c - γ etc.) ⇒ y = c - x ⇒ x = c - y
- M1 (2 – y)³ – 2(2 – y)² – 3(2 – y) + 4 = 0 ...(their c)
- A1 to obtain required cubic equation. ⇒ y³ – 4y² + y + 2 = 0
- M1A1 Obtains equation whose roots are reciprocals of those in previous cubic equation. Uses z = y⁻¹ to obtain 2z³ + z² – 4z + 1 = 0
- M1A1 Uses Σα² = (Σα)² – 2Σαβ and Σ(1/(α+β))² = ((-1/2))² – 2(-2/2) = 1/4 + 2 = 9/4
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