Question 7 - A-Level Maths

OCR MEI Further Pure Core AS 2020 November — Question 7 7 marks

Exam Board	OCR MEI
Module	Further Pure Core AS (Further Pure Core AS)
Year	2020
Session	November
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Roots of polynomials
Type	Complex roots with real coefficients
Difficulty	Standard +0.8 This is a Further Maths question requiring knowledge that complex roots come in conjugate pairs for real coefficients, then using sum/product of roots to find remaining roots. It involves multiple steps (identifying conjugate pair, using Vieta's formulas, solving resulting quadratic) but follows a standard Further Maths technique without requiring novel insight.
Spec	4.02g Conjugate pairs: real coefficient polynomials

7 In the quartic equation \(2 x ^ { 4 } - 20 x ^ { 3 } + a x ^ { 2 } + b x + 250 = 0\), the coefficients \(a\) and \(b\) are real. One root of the equation is \(2 + \mathrm { i }\). Find the other roots.

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Question 7:

Answer	Marks
7	Another root is 2 – i

suppose other roots are γ, δ (say)

4 + γ + δ = 10

⇒ γ + δ = 6

(2+i)(2–i)γδ = 125

5γδ = 125

γ = 3 + 4i, δ = 3 − 4i

Answer	Marks
So other 2 roots are 3 + 4i and 3 − 4i	B1

M1

A1

M1

A1

M1

Answer	Marks
A1	1.2

3.1a

1.1a

1.1

Answer	Marks
3.2a	product of roots used
eliminating and solving quad	or c + di and c − di. M1

4+2c = 10 M1

⇒ c = 3 A1

(2+i)(2–i)(3+di)(3–di) = 125

⇒ 5(9 + d2) = 125 M1A1

⇒ d = 4 A1

Answer	Marks
Alternative solution	B1

M1

A1

M1

A1

M1

Answer	Marks
A1	Attempt to multiply factors

x2 – 4x + 5

Attempt to find other quad

factor

Answer	Marks
2x2 − 12x + 50	condone sign errors

or long division

Another root is 2 − i

(x − 2 − i)(x − 2 + i) = x2 – 4x + 5

(x2–4x+5)(2x2 + kx +50) = 2x4−20x3+ax2+bx+250

⇒ k − 8 = −20, so k = −12

Other factor is 2x2 – 12x + 50

x2 –6x + 25 =0 gives x = 3 ± 4i

Roots are 2 + i, 2 – i, 3 + 4i and 3 – 4i

[7]

Question 7:
7 | Another root is 2 – i
suppose other roots are γ, δ (say)
4 + γ + δ = 10
⇒ γ + δ = 6
(2+i)(2–i)γδ = 125
5γδ = 125
γ = 3 + 4i, δ = 3 − 4i
So other 2 roots are 3 + 4i and 3 − 4i | B1
M1
A1
M1
A1
M1
A1 | 1.2
3.1a
1.1a
1.1
1.1
1.1
3.2a | product of roots used
eliminating and solving quad | or c + di and c − di. M1
4+2c = 10 M1
⇒ c = 3 A1
(2+i)(2–i)(3+di)(3–di) = 125
⇒ 5(9 + d2) = 125 M1A1
⇒ d = 4 A1
Alternative solution | B1
M1
A1
M1
A1
M1
A1 | Attempt to multiply factors
x2 – 4x + 5
Attempt to find other quad
factor
2x2 − 12x + 50 | condone sign errors
or long division
Another root is 2 − i
(x − 2 − i)(x − 2 + i) = x2 – 4x + 5
(x2–4x+5)(2x2 + kx +50) = 2x4−20x3+ax2+bx+250
⇒ k − 8 = −20, so k = −12
Other factor is 2x2 – 12x + 50
x2 –6x + 25 =0 gives x = 3 ± 4i
Roots are 2 + i, 2 – i, 3 + 4i and 3 – 4i
[7]

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7 In the quartic equation $2 x ^ { 4 } - 20 x ^ { 3 } + a x ^ { 2 } + b x + 250 = 0$, the coefficients $a$ and $b$ are real. One root of the equation is $2 + \mathrm { i }$.

Find the other roots.

\hfill \mbox{\textit{OCR MEI Further Pure Core AS 2020 Q7 [7]}}

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