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Byju's Answer
Standard IX
Mathematics
Zeroes of a Polynomial
Solve the pro...
Question
Solve the problem
I
f
α
,
β
,
γ
a
r
e
t
h
e
r
o
o
t
s
o
f
t
h
e
e
q
u
a
t
i
o
n
x
3
-
3
x
+
11
=
0
t
h
e
n
f
i
n
d
t
h
e
e
q
u
a
t
i
o
n
w
h
o
s
e
r
o
o
t
s
a
r
e
α
+
β
,
β
+
γ
a
n
d
γ
+
α
.
Open in App
Solution
Dear Student,
x
3
-
3
x
+
11
=
0
Comparing
with
ax
3
+
bx
2
+
cx
+
d
=
0
Here
,
a
=
1
b
=
0
c
=
-
3
d
=
11
We
know
that
for
a
cubic
equation
,
α
+
β
+
γ
=
-
b
a
=
-
0
1
=
0
.
.
.
(
i
)
and
,
αβ
+
βγ
+
γα
=
c
a
=
-
3
1
=
-
3
and
,
αβγ
=
-
d
a
=
-
11
1
=
-
11
Now
,
if
there
is
a
cubic
polynomial
of
roots
α
+
β
,
β
+
γ
and
γ
+
α
,
then
it
zeroes
will
be
-
γ
,
-
α
and
-
β
respectively
(
from
eq
i
)
Sum
of
zeroes
=
(
-
γ
)
+
(
-
α
)
+
(
-
β
)
=
-
(
γ
+
α
+
β
)
=
0
Sum
of
product
of
its
two
zeroes
=
(
-
α
)
(
-
β
)
+
(
-
β
)
(
-
γ
)
+
(
-
γ
)
(
-
α
)
or
,
αβ
+
βγ
+
γα
=
-
3
and
,
Product
of
its
zeroes
=
(
-
α
)
(
-
β
)
(
-
γ
)
=
-
αβγ
=
-
(
-
11
)
=
11
So
,
required
polynomial
=
x
3
-
(
Sum
of
zeroes
)
x
2
+
(
Sum
of
product
of
its
two
zeroes
)
x
-
(
Product
of
its
zeroes
)
=
x
3
-
0
x
2
-
3
x
-
11
=
x
3
-
3
x
-
11
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0
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β
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α
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