Question 13:
--- 13(a)(i) ---
13(a)(i) | Recognises that result
can be obtained by
(α+β+γ)2
expanding
and completes correct
expansion. | AO3.1a | M1 | (α+β+γ)2 =
α2+β2+γ2+2(αβ+βγ+γα)
α+β+γ=−k
But and
αβ+βγ+γα=0
So
α2 +β2 +γ2 =k2
Correctly uses sum of
roots and sum of pairs of
roots, in their expansion.
Condone α+β+γ=k | AO1.1a | B1
Completes a rigorous
argument to show the
required result.
Do not condone
α+β+γ=k | AO2.1 | R1
--- 13(a)(ii) ---
13(a)(ii) | (αβ+βγ+γα)2
Expands | AO1.1a | M1 | (αβ+βγ+γα)2 =α2β2+β2γ2+γ2α2
+2αβ2γ+2α2βγ+2αβγ2
=α2β2+β2γ2+γ2α2
+2αβγ(α+β+γ)
α+β+γ=−k
But
αβ+βγ+γα=0
And
αβγ=−9
And
So α2β2 +β2γ2 +γ2α2 =−18k
Rearranges to express
α2β2 +β2γ2 +γ2α2
in
terms of sum, product
and sum of pairs of roots | AO3.1a | M1
Correctly states product
of roots – could be seen
in (a)(i) | AO1.1a | B1
Completes a rigorous
argument to show the
required result. | AO2.1 | R1
--- 13(b)(i) ---
13(b)(i) | Deduces result correctly
using both equations and
previous working. | AO2.2a | R1 | 40
αβ+γ+βγ+α+γα+β=
9
α+β+γ=−k αβ+βγ+γα=0
But And
40
so k =−
9
--- 13(b)(ii) ---
13(b)(ii) | Forms correct equation
from product of roots | AO1.1b | B1 | s
− =(αβ+γ)(βγ+α)(γα+β)
9
=α2β2γ2 +αβ3γ+α3βγ+α2β2
+αβγ3+β2γ2 +γ2α2 +αβγ
=(αβγ)2 +αβγ+α2β2 +β2γ2 +γ2α2
( )
+αβγ α2 +β2 +γ2
=81−9−18k−9k2
1600
=81−9+80−9×
81
232
=−
9
s=232
So
Expands product of
roots. Condone one or
two errors | AO1.1a | M1
Identifies the degree 5
terms and fully factorises
them in the form
( α2+β2+γ2)
αβγ | AO3.1a | M1
Collects all other terms
on RHS as
(αβγ)2+αβγ+α2β2+β2γ2+γ2α2
(αβγ)2 PI 81 or (−9 )2 | AO3.1a | M1
Correctly substitutes
their values into their
equation | AO1.1a | M1
Obtains the correct
answer from correct
reasoning. | AO2.1 | R1
Total | 14
Q | Marking Instructions | AO | Marks | Typical solution