IBDP Maths AHL 1.14 Conjugate roots of polynomial equations AA HL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a) Since the polynomial has real coefficients, complex roots occur in conjugate pairs. Given roots \( c + i \) and \( 2 + id \), the other two roots are their conjugates:
\[ c – i \text{ and } 2 – id \] A1
[1 mark]
(b) Use the sum and product of roots for the quartic polynomial \( z^4 – 10z^3 + az^2 + bz + 50 = 0 \).
Sum of roots: \( (c + i) + (c – i) + (2 + id) + (2 – id) = c + c + 2 + 2 = 2c + 4 \).
For a quartic \( z^4 + pz^3 + qz^2 + rz + s \), the sum of roots is \( -p \). Here, \( p = -10 \):
\[ 2c + 4 = 10 \] M1
\[ 2c = 6 \implies c = 3 \] A1
Product of roots: \( (c + i)(c – i)(2 + id)(2 – id) = (c^2 + 1)(4 + d^2) \). M1 A1
For the polynomial, the product of roots is the constant term: \( 50 \).
\[ (c^2 + 1)(4 + d^2) = 50 \]
Substitute \( c = 3 \):
\[ (3^2 + 1)(4 + d^2) = 10(4 + d^2) = 50 \] A1
\[ 4 + d^2 = 5 \implies d^2 = 1 \implies d = 1 \quad (\text{since } d > 0) \] A1
Thus, \( c = 3 \), \( d = 1 \).
[6 marks]
Total [7 marks]
▶️ Answer/Explanation
(a) Since the polynomial has real coefficients, complex roots occur in conjugate pairs. Given the root \( -1 + 3i \), its conjugate is also a root:
\[ -1 – 3i \] A1
The polynomial is cubic, so there is one more root. The points \( -1 + 3i \), \( -1 – 3i \), and the third root form a triangle with area 9 in the Argand diagram.
Distance between \( -1 + 3i \) and \( -1 – 3i \): \( |(-1 + 3i) – (-1 – 3i)| = |6i| = 6 \). This is the base of the triangle. M1
Area of a triangle is \( \frac{1}{2} \cdot \text{base} \cdot \text{height} = 9 \):
\[ \frac{1}{2} \cdot 6 \cdot \text{height} = 9 \implies \text{height} = 3 \] A1
The third root has real part \( x \), and the perpendicular distance from \( x \) to the line joining \( -1 + 3i \) and \( -1 – 3i \) (at \( x = -1 \)) is the height. Thus, \( |x – (-1)| = 3 \implies x = -1 \pm 3 \).
Possible third roots: \( x = 2 \) or \( x = -4 \). A1
Thus, the other two roots are \( -1 – 3i \) and either \( 2 \) or \( -4 \).
[4 marks]
(b) Consider both possible third roots to find \( a, b, c \).
Case 1: Third root is 2
Roots: \( -1 + 3i, -1 – 3i, 2 \). Form the polynomial:
\[ (z – (-1 + 3i))(z – (-1 – 3i))(z – 2) \] M1
\[ (z + 1 – 3i)(z + 1 + 3i) = z^2 + (1 – 3i + 1 + 3i)z + (1 – 3i)(1 + 3i) = z^2 + 2z + (1 + 9) = z^2 + 2z + 10 \] A1
\[ (z^2 + 2z + 10)(z – 2) = z^3 – 2z^2 + 2z^2 – 4z + 10z – 20 = z^3 + 6z – 20 \] A1
Compare with \( z^3 + az^2 + bz + c \):
\[ a = 0, b = 6, c = -20 \]
Case 2: Third root is -4
Roots: \( -1 + 3i, -1 – 3i, -4 \). Form the polynomial:
\[ (z – (-1 + 3i))(z – (-1 – 3i))(z – (-4)) \] M1
\[ (z^2 + 2z + 10)(z + 4) = z^3 + 4z^2 + 2z^2 + 8z + 10z + 40 = z^3 + 6z^2 + 18z + 40 \] A1
Compare with \( z^3 + az^2 + bz + c \):
\[ a = 6, b = 18, c = 40 \] A1
Thus, the possible values are \( (a, b, c) = (0, 6, -20) \) or \( (6, 18, 40) \).
[3 marks]
Total [7 marks]