IB Mathematics AHL 1.14 Conjugate roots of polynomial equations AA HL Paper 3
▶️ Answer/Explanation
(a) Show coefficients of cubic:
Expand \( (x – \alpha)(x – \beta)(x – \gamma) \):
\[ (M1) \]
\[ = \left( x^2 – (\alpha + \beta) x + \alpha \beta \right)(x – \gamma) \]
Or: \( (x – \alpha) \left( x^2 – (\beta + \gamma) x + \beta \gamma \right) \).
\[ = x^3 – (\alpha + \beta + \gamma) x^2 + (\alpha \beta + \beta \gamma + \gamma \alpha) x – \alpha \beta \gamma \]
Compare coefficients with \( x^3 + p x^2 + q x + r \):
\[ p = -(\alpha + \beta + \gamma) \quad (A1)(AG) \]
\[ q = \alpha \beta + \beta \gamma + \gamma \alpha \quad (A1)(AG) \]
\[ r = -\alpha \beta \gamma \quad (A1)(AG) \]
Note: For candidates who do not include the (AG) lines, award full marks.
[3 marks]
(b) (i) Show \( p^2 – 2q = \alpha^2 + \beta^2 + \gamma^2 \):
\[ p^2 – 2q = (\alpha + \beta + \gamma)^2 – 2(\alpha \beta + \beta \gamma + \gamma \alpha) \]
Expand \( (\alpha + \beta + \gamma)^2 \):
\[ (M1) \]
\[ = \alpha^2 + \beta^2 + \gamma^2 + 2(\alpha \beta + \beta \gamma + \gamma \alpha) – 2(\alpha \beta + \beta \gamma + \gamma \alpha) \]
\[ = \alpha^2 + \beta^2 + \gamma^2 \quad (A1)(AG) \]
Note: Accept equivalent working from RHS to LHS.
[3 marks]
(b) (ii) Show \( (\alpha – \beta)^2 + (\beta – \gamma)^2 + (\gamma – \alpha)^2 = 2p^2 – 6q \):
Either:
Expand \( (\alpha – \beta)^2 + (\beta – \gamma)^2 + (\gamma – \alpha)^2 \):
\[ (M1) \]
\[ = (\alpha^2 + \beta^2 – 2 \alpha \beta) + (\beta^2 + \gamma^2 – 2 \beta \gamma) + (\gamma^2 + \alpha^2 – 2 \gamma \alpha) \]
\[ = 2(\alpha^2 + \beta^2 + \gamma^2) – 2(\alpha \beta + \beta \gamma + \gamma \alpha) \]
\[ = 2(p^2 – 2q) – 2q \]
\[ = 2p^2 – 6q \quad (A1)(AG) \]
Or:
Write \( 2p^2 – 6q \) in terms of \( \alpha, \beta, \gamma \):
\[ (M1) \]
\[ = 2(p^2 – 2q) – 2q \]
\[ = 2(\alpha^2 + \beta^2 + \gamma^2) – 2(\alpha \beta + \beta \gamma + \gamma \alpha) \]
\[ = (\alpha^2 + \beta^2 – 2 \alpha \beta) + (\beta^2 + \gamma^2 – 2 \beta \gamma) + (\gamma^2 + \alpha^2 – 2 \gamma \alpha) \]
\[ = (\alpha – \beta)^2 + (\beta – \gamma)^2 + (\gamma – \alpha)^2 \quad (A1)(AG) \]
Note: Accept equivalent working where LHS and RHS are expanded to identical expressions.
[3 marks]
(c) Deduce roots cannot all be real if \( p^2 < 3q \):
\[ p^2 < 3q \implies 2p^2 – 6q < 0 \]
\[ \implies (\alpha – \beta)^2 + (\beta – \gamma)^2 + (\gamma – \alpha)^2 < 0 \quad (A1) \]
If all roots are real, \( (\alpha – \beta)^2 + (\beta – \gamma)^2 + (\gamma – \alpha)^2 \geq 0 \).
\[ (R1) \]
Note: Condone strict inequality in the (R1) line.
Thus, roots cannot all be real \quad (AG).
Note: Do not award (A0)(R1).
[2 marks]
(d) Show \( q = 17 \) implies at least one complex root:
For \( x^3 – 7 x^2 + q x + 1 = 0 \), \( p = -7 \):
\[ p^2 = (-7)^2 = 49 \]
\[ q = 17, \quad 3q = 3 \cdot 17 = 51 \quad (A1) \]
\[ p^2 < 3q \quad (49 < 51) \]
By part (c), the equation has at least one complex root \quad (R1).
Note: Allow equivalent comparisons, e.g., checking \( p^2 < 6q \).
[2 marks]
(e) (i) Smallest positive integer \( q \) to show Noah is incorrect:
Use GDC (e.g., graphs or tables) to find \( q \) where \( p^2 \geq 3q \) but not all roots are real:
\[ (M1) \]
\[ q = 12 \quad (A1) \]
[2 marks]
(e) (ii) Explain at least one real root:
Complex roots appear in conjugate pairs, so if complex roots occur, the other root is real, or all three roots are real.
Or: A cubic curve always crosses the x-axis at least once \quad (R1).
[1 mark]
(f) (i) Expression for \( \alpha^2 + \beta^2 + \gamma^2 + \delta^2 \):
Expand \( (\alpha + \beta + \gamma + \delta)^2 \):
\[ (M1) \]
\[ = \alpha^2 + \beta^2 + \gamma^2 + \delta^2 + 2(\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta) \]
\[ \alpha^2 + \beta^2 + \gamma^2 + \delta^2 = (\alpha + \beta + \gamma + \delta)^2 – 2(\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta) \]
\[ = p^2 – 2q \quad (A1) \]
[3 marks]
(f) (ii) Condition for at least one complex root:
\[ p^2 < 2q \quad \text{or} \quad p^2 – 2q < 0 \quad (A1) \]
Note: Allow FT on their result from part (f)(i).
[1 mark]
(g) Show \( x^4 – 2 x^3 + 3 x^2 – 4 x + 5 = 0 \) has at least one complex root:
For \( p = -2 \), \( q = 3 \):
\[ p^2 = 4, \quad 2q = 6 \]
\[ 4 < 6 \quad \text{or} \quad p^2 – 2q = 4 – 6 = -2 < 0 \]
\[ (R1) \]
Hence, there is at least one complex root \quad (AG).
Note: Allow FT from part (f)(ii) for the (R) mark provided numerical reasoning is seen.
[1 mark]
(h) (i) Result of part (f)(ii) for \( x^4 – 9 x^3 + 24 x^2 + 22 x – 12 = 0 \):
\[ p = -9, \quad q = 24 \]
\[ p^2 = 81, \quad 2q = 2 \cdot 24 = 48 \]
\[ 81 > 48 \quad (p^2 > 2q), \text{ so nothing can be deduced} \]
\[ (R1) \]
Note: Do not allow FT for the (R) mark.
[1 mark]
(h) (ii) Integer root:
\[ -1 \quad (A1) \]
[1 mark]
(h) (iii) Prove at least one complex root:
Attempt to express as a product of linear and cubic factors:
\[ (M1) \]
\[ x^4 – 9 x^3 + 24 x^2 + 22 x – 12 = (x + 1)(x^3 – 10 x^2 + 34 x – 12) \]
\[ (A1)(A1) \]
Note: Award (A1) for each factor. Award at most (A1)(A0) if not written as a product.
For the cubic \( x^3 – 10 x^2 + 34 x – 12 \), check \( p^2 < 3q \):
\[ p = -10, \quad q = 34 \]
\[ p^2 = 100, \quad 3q = 3 \cdot 34 = 102 \]
\[ 100 < 102 \quad (R1) \]
By part (c), the cubic has at least one complex root \quad (AG).
[4 marks]