In this question you must show detailed reasoning.
The cubic equation \(5x^3 + 3x^2 - 4x + 7 = 0\) has roots \(\alpha\), \(\beta\) and \(\gamma\).
Find a cubic equation with integer coefficients whose roots are \(\alpha + \beta\), \(\beta + \gamma\) and \(\gamma + \alpha\). [7]
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Question 5:
α+β+γ=− , αβ+βγ+γα=− , αβγ=−
5 5 5
6
Α + Β + Γ = 2(α + β + γ) = −
5
ΑΒ + ΒΓ + ΓΑ = (α + β)(β + γ) + (β + γ)(γ + α) +
(γ + α)(α + β) = 3(αβ + βγ + γα) + α2 + β2 + γ2
11
= αβ + βγ + γα + (α + β + γ)2 = −
25
ΑΒΓ = (α + β)(β + γ)(γ + α) = 2αβγ + α(αβ + βγ
+ γα) – αβγ + β(αβ + βγ + γα) – αβγ + γ (αβ + βγ
+ γα) – αβγ = (α + β + γ)(αβ + βγ + γα) – αβγ
3 4 7 12 35 47
=− ×− −− = + =
5 5 5 25 25 25
Answer Marks
a = 25 => 25x3 + 30x2 – 11x – 47 = 0 B1
B1ft
M1
A1
M1
A1
2.1
2.1
1.1
2.1
1.1
Answer Marks
1.1 For at least 2 correct
2× their Σα
Attempting to find the new Σαβ and
expanding to a symmetrical form
Opening brackets convincingly and
writing in symmetrical form
Answer Marks
Or any non-zero integer multiple Α, Β and Γ are the roots of the new
equation
Other correct forms are possible eg
2 2 2 2 2
𝛼𝛼 𝛽𝛽+𝛼𝛼 𝛾𝛾+𝛽𝛽 𝛼𝛼+𝛽𝛽 𝛾𝛾+𝛾𝛾 𝛼𝛼
2
+𝛾𝛾 𝛽𝛽+2𝛼𝛼𝛽𝛽𝛾𝛾
Must be integer coefficients
Alternative method
3
Substitution x=− −u
B1dep*
Answer Marks
B1* Used or stated
3 4 7
α+β+γ=− , αβ+βγ+γα=− , αβγ=−
5 5 5
Answer Marks
Guidance
B1dep* Might occur before the first B1 in
Note only the sum of roots needed
candidates working here
B1** Show that when x is one of the roots of
the original equation, u is one of the
roots of the new equation
When x=α,
3
u = − −α
5
= α+β+γ−α
Answer Marks
= α+β Show that when x is one of the roots of
the original equation, u is one of the
roots of the new equation
B1
dep**
Show that x=− −u gives the other
5
two required roots
3
Use substitution x=− −u and
5
3 3 3 2
expand − −u and − −u
Answer Marks
5 5 3 3 9 27 27
− −u =−u3+ u2+ u+
5 5 25 125
When x=β/γ,
3
u = − −β/γ
5
= α+β+γ−β/γ
= α+γ/β
3 3 3 2 3
5− −u +3− −u −4− −u+7=0
5 5 5
( u3+3u2⋅3+3u⋅(3)2+(3)3)
−5
5 5 5
( u2+2u⋅3+(3)2)
+3 +4 ( u+3)+7=0
5 5 5
11 47
−5u3−6u2+ u+ [=0]
5 5
Answer Marks
25u3+30u2−11u−47=0 11 47
−5u3−6u2+ u+ [=0]
Answer Marks
5 5 A1
A1 Or any non-zero integer multiple
25u3+30u2−11u−47=0
Must be integer coefficients. Must
have “=0”
[7]
B1**
B1
dep**
M1
3
Show that x=− −u gives the other
5
two required roots
3
Use substitution x=− −u and
5
3 3 3 2
expand − −u and − −u
5 5
3 3 9 27 27
− −u =−u3+ u2+ u+
5 5 25 125
A1
A1
Or any non-zero integer multiple
Copy
Question 5:
5 | 3 4 7
α+β+γ=− , αβ+βγ+γα=− , αβγ=−
5 5 5
6
Α + Β + Γ = 2(α + β + γ) = −
5
ΑΒ + ΒΓ + ΓΑ = (α + β)(β + γ) + (β + γ)(γ + α) +
(γ + α)(α + β) = 3(αβ + βγ + γα) + α2 + β2 + γ2
11
= αβ + βγ + γα + (α + β + γ)2 = −
25
ΑΒΓ = (α + β)(β + γ)(γ + α) = 2αβγ + α(αβ + βγ
+ γα) – αβγ + β(αβ + βγ + γα) – αβγ + γ (αβ + βγ
+ γα) – αβγ = (α + β + γ)(αβ + βγ + γα) – αβγ
3 4 7 12 35 47
=− ×− −− = + =
5 5 5 25 25 25
a = 25 => 25x3 + 30x2 – 11x – 47 = 0 | B1
B1ft
M1
A1
M1
A1
A1 | 1.1a
2.1
2.1
1.1
2.1
1.1
1.1 | For at least 2 correct
2× their Σα
Attempting to find the new Σαβ and
expanding to a symmetrical form
Opening brackets convincingly and
writing in symmetrical form
Or any non-zero integer multiple | Α, Β and Γ are the roots of the new
equation
Other correct forms are possible eg
2 2 2 2 2
𝛼𝛼 𝛽𝛽+𝛼𝛼 𝛾𝛾+𝛽𝛽 𝛼𝛼+𝛽𝛽 𝛾𝛾+𝛾𝛾 𝛼𝛼
2
+𝛾𝛾 𝛽𝛽+2𝛼𝛼𝛽𝛽𝛾𝛾
Must be integer coefficients
Alternative method
3
Substitution x=− −u
5 | B1*
B1dep*
B1* | Used or stated
3 4 7
α+β+γ=− , αβ+βγ+γα=− , αβγ=−
5 5 5
B1dep* | Might occur before the first B1 in | Note only the sum of roots needed
candidates working | here
B1** | Show that when x is one of the roots of
the original equation, u is one of the
roots of the new equation
When x=α,
3
u = − −α
5
= α+β+γ−α
= α+β | Show that when x is one of the roots of
the original equation, u is one of the
roots of the new equation
B1
dep**
M1 | 3
Show that x=− −u gives the other
5
two required roots
3
Use substitution x=− −u and
5
3 3 3 2
expand − −u and − −u
5 5 | 3 3 9 27 27
− −u =−u3+ u2+ u+
5 5 25 125
When x=β/γ,
3
u = − −β/γ
5
= α+β+γ−β/γ
= α+γ/β
3 3 3 2 3
5− −u +3− −u −4− −u+7=0
5 5 5
( u3+3u2⋅3+3u⋅(3)2+(3)3)
−5
5 5 5
( u2+2u⋅3+(3)2)
+3 +4 ( u+3)+7=0
5 5 5
11 47
−5u3−6u2+ u+ [=0]
5 5
25u3+30u2−11u−47=0 | 11 47
−5u3−6u2+ u+ [=0]
5 5 | A1
A1 | Or any non-zero integer multiple
25u3+30u2−11u−47=0
Must be integer coefficients. Must
have “=0”
[7]
B1**
B1
dep**
M1
3
Show that x=− −u gives the other
5
two required roots
3
Use substitution x=− −u and
5
3 3 3 2
expand − −u and − −u
5 5
3 3 9 27 27
− −u =−u3+ u2+ u+
5 5 25 125
A1
A1
Or any non-zero integer multiple
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In this question you must show detailed reasoning.
The cubic equation $5x^3 + 3x^2 - 4x + 7 = 0$ has roots $\alpha$, $\beta$ and $\gamma$.
Find a cubic equation with integer coefficients whose roots are $\alpha + \beta$, $\beta + \gamma$ and $\gamma + \alpha$. [7]
\hfill \mbox{\textit{OCR Further Pure Core AS 2020 Q5 [7]}}