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Byju's Answer
Standard X
Mathematics
Nature of Roots
If the roots ...
Question
If the roots of the equation
1
x
+
p
+
1
x
+
q
=
1
r
are equal in magnitude and opposite in sign, then
A
p
+
q
=
r
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B
p
+
q
=
2
r
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C
Product of roots
=
−
1
2
(
p
2
+
q
2
)
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D
Sum of roots
=
1
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Solution
The correct options are
B
p
+
q
=
2
r
C
Product of roots
=
−
1
2
(
p
2
+
q
2
)
Given equation is :
1
x
+
p
+
1
x
+
q
=
1
r
⇒
r
(
2
x
+
p
+
q
)
=
(
x
+
p
)
(
x
+
q
)
⇒
x
2
+
(
p
+
q
−
2
r
)
x
+
p
q
−
p
r
−
q
r
=
0
…
(
1
)
From the equation we get, sum of roots
=
−
(
p
+
q
−
2
r
)
1
=
2
r
−
p
−
q
Product of roots
=
p
q
−
p
r
−
q
r
1
=
p
q
−
p
r
−
q
r
As the roots are equal and opposite,we get sum of roots
=
0
⇒
2
r
−
p
−
q
=
0
⇒
p
+
q
=
2
r
…
(
2
)
Substituting
r
from equation
(
2
)
in equation
(
1
)
, we get Product of roots
=
p
q
−
p
(
p
+
q
2
)
−
q
(
p
+
q
2
)
=
p
q
−
p
2
2
−
p
q
2
−
q
p
2
−
q
2
2
=
−
p
2
2
−
q
2
2
=
−
1
2
(
p
2
+
q
2
)
…
(
3
)
Hence, options (B) and (C) are the correct choices.
Suggest Corrections
0
Similar questions
Q.
If the roots of the equation
1
(
x
+
p
)
+
1
(
x
+
q
)
=
1
r
are equal in magnitude but opposite in sign, show that
p
+
q
=
2
r
and that the product of the roots is equal to
−
1
2
(
p
2
+
q
2
)
.
Q.
Let
p
,
q
a
n
d
r
be real numbers
(
p
≠
q
,
r
≠
0
)
, such that the roots of the equation
1
x
+
p
+
1
x
+
q
=
1
r
are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to :
Q.
Let p, q, and r be real numbers
(
p
≠
q
,
r
≠
0
)
, such that the roots of the equation
1
x
+
p
+
1
x
+
q
=
1
r
are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to.
Q.
If the roots of the equation
1
x
+
p
+
1
x
+
q
=
1
r
are equal in magnitude but opposite in sign, then the product of the roots is :
Q.
If the roots of
1
x
+
p
+
1
x
+
q
=
1
r
are equal in magnitude and opposite in sign, then product of
roots is
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