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Byju's Answer
Standard X
Mathematics
Solving QE Using Quadratic Formula When D>0
If the roots ...
Question
If the roots of the equation
(
x
−
a
)
(
x
−
b
)
−
k
=
0
be
c
and
d
,
then prove that the roots of the equation
(
x
−
c
)
(
x
−
d
)
+
k
=
0
are
a
and
b
.
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Solution
According to the question,
(
x
−
a
)
(
x
−
b
)
−
k
≡
(
x
−
c
)
(
x
−
d
)
or
(
x
−
c
)
(
x
−
d
)
+
k
≡
(
x
−
a
)
(
x
−
b
)
This relation shows that the roots of
(
x
−
c
)
(
x
−
d
)
+
k
=
0
are
a
and
b
Hence, Proved.
Suggest Corrections
0
Similar questions
Q.
Let
a
,
b
,
c
and
d
be non-zero numbers such that
x
=
c
and
x
=
d
are the roots of the equation
x
2
+
a
x
+
b
=
0
and
x
=
a
and
x
=
b
are the roots of the equation
x
2
+
c
x
+
d
=
0.
Then
Q.
If one root of the equation
x
2
+
a
x
+
b
=
0
is a root of the equation
x
2
+
c
x
+
d
=
0
, prove that the other roots satisfy the equation
x
2
+
x
(
2
a
−
c
)
+
(
a
2
−
a
c
+
d
)
=
0
.
Q.
If the roots of
a
x
3
+
b
x
2
+
c
x
+
d
=
0
are in A.P., then the roots of
a
(
x
+
k
)
3
+
b
(
x
+
k
)
2
+
c
(
x
+
k
)
+
d
=
0
are in
Q.
Assertion :If
a
+
b
+
c
>
0
,
a
<
0
<
b
<
c
, then roots of the equation
a
(
x
−
b
)
(
x
−
c
)
+
b
(
x
−
c
)
(
x
−
a
)
+
c
(
x
−
a
)
(
x
−
b
)
=
0
are real. Reason: Roots of the equation
A
x
2
+
B
x
+
K
=
0
are real if
B
2
−
4
A
K
>
0
.
Q.
If (2+i) and
(
√
5
−
2
i
)
are the roots of the. equation
(
x
2
+
a
x
−
+
b
)
(
x
2
+
c
x
+
d
)
=
0
, where
a
,
b
,
c
and d are real constants, then product of all the roots of the equation is
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