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Past exam of the mathematics course of the University of Cambridge
/
2025
/
ii
/
Paper 1
/
18J
/
a
/
Solution
...
Past exam of the mathematics course of the University of Cambridge
2025
ii
Paper 1
18J
a
OurBigBook.com
Words: 59
For a monic
polynomial
with roots
r
1
,
…
,
r
n
, its discriminant is
Disc
(
f
)
=
∏
i
<
j
(
r
i
−
r
j
)
2
.
(94)
Here
Disc
(
g
)
=
(
u
3
−
v
3
)
2
. Both
u
3
−
v
3
and the Vandermonde product in the
α
i
are alternating homogeneous cubics. Evaluation at
(
α
1
,
α
2
,
α
3
)
=
(
0
,
1
,
−
1
)
fixes the constant and gives
(
u
3
−
v
3
)
2
=
−
27
∏
i
<
j
(
α
i
−
α
j
)
2
=
−
27
Disc
(
f
)
.
(95)
Using
α
1
+
α
2
+
α
3
=
0
,
∑
i
<
j
α
i
α
j
=
a
, and
α
1
α
2
α
3
=
−
b
, direct expansion gives
uv
=
−
3
a
,
u
3
+
v
3
=
−
27
b
.
(96)
Consequently
g
(
X
)
=
X
2
+
27
b
X
−
27
a
3
,
(97)
and
Disc
(
g
)
=
729
b
2
+
108
a
3
=
−
27
(
−
4
a
3
−
27
b
2
)
.
(98)
Thus
Disc
(
f
)
=
−
4
a
3
−
27
b
2
.
(99)
Solved by gpt-5.6-sol high.
Ancestors
(11)
A
18J
Paper 1
Ii
2025
Past exam of the mathematics course of the University of Cambridge
Mathematics course of the University of Cambridge
Course of the University of Cambridge
University of Cambridge
List of universities
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