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Byju's Answer
Standard XII
Mathematics
Modulus of a Complex Number
Let x1, x2 ...
Question
Let
x
1
,
x
2
are the roots of the quadratic equation
x
2
+
a
x
+
b
=
0
where
a
,
b
are complex numbers and
y
1
,
y
2
are the roots of the quadratic equation
y
2
+
∣
a
∣
y
+
∣
b
∣
=
0
. If
∣
x
1
∣
=
∣
x
2
∣
=
1
then prove that
∣
y
1
∣
=
∣
y
2
∣
=
1
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Solution
Suppose
x
2
+
a
x
+
b
=
0
has root
x
1
x
2
. Then,
∴
x
1
+
x
2
=
−
a
and
x
1
x
2
=
b
From (2),
|
x
1
|
|
x
2
|
=
|
b
|
⇒
|
b
|
=
1
Also,
|
−
a
|
=
|
x
1
+
x
2
|
⇒
|
a
|
≤
|
x
1
|
+
|
x
2
|
or
|
a
|
≤
2
Now suppose
y
2
+
|
a
|
y
+
|
b
|
=
0
has roots
y
1
a
n
d
y
2
Then,
y
1
=
y
2
=
−
|
a
|
±
√
|
a
|
2
−
4
|
b
|
2
=
−
|
a
|
±
(
√
4
−
|
a
|
2
)
i
2
⇒
|
y
1
|
=
|
y
2
|
=
√
|
a
|
2
+
4
−
|
a
|
2
2
=
1
Hence,
|
y
1
|
=
|
y
2
|
=
1
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0
Similar questions
Q.
Let
x
1
,
x
2
,
are the roots of quadratic equation
x
2
+
a
x
+
b
=
0
, Where
a
,
b
are complex numbers and
y
1
,
y
2
are the roots of the quadratic equation
y
2
+
|
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|
y
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|
b
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=
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x
1
|
=
|
x
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|
=
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Q.
Q
30
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t
h
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p
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n
t
s
(
x
1
,
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1
)
,
(
x
2
,
y
2
)
a
n
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Q.
Let
P
(
x
1
,
y
1
)
and
Q
(
x
2
,
y
2
)
are two points such that their abscissa
x
1
and
x
2
are the roots of the equation
x
2
+
2
x
−
3
=
0
while the ordinates
y
1
and
y
2
are the roots of the equation
y
2
+
4
y
−
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=
0.
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The equation of the chord joining two points
(
x
1
,
y
1
)
and
(
x
2
,
y
2
)
on the rectangular hyperbola
x
y
=
c
2
is
Q.
If
x
1
,
y
1
are the roots of
x
2
+
8
x
−
20
=
0
and
x
2
,
y
2
are the roots of
4
x
2
+
32
x
−
57
=
0
and
x
3
,
y
3
are the roots of
9
x
2
+
72
x
−
112
=
0
such that
y
i
<
0
,
then the points
(
x
1
,
y
1
)
,
(
x
2
,
y
2
)
and
(
x
3
,
y
3
)
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