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JEE Main Math Class 11 Complex Numbers and Quadratic Equations Part 1 Questions
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© examsnet.com
Question : 2
Total: 100
If one real root of the quadratic equation
81
x
2
+
k
x
+
256
=
0
is cube of the other root, then a value of
k
is :
[Jan. 11, 2019 (I)]
-81
100
144
-300
Validate
Solution:
Let
α
and
β
be the roots of the equation,
81
x
2
+
k
x
+
256
=
0
Given
(
α
)
1
3
=
β
⇒
α
=
β
3
∵
Product of the roots
=
256
81
∴
(
α
)
(
β
)
=
256
81
⇒
β
4
=
(
4
3
)
4
⇒
β
=
4
3
⇒
α
=
64
27
Sum of the roots
=
−
k
81
∴
α
+
β
=
−
k
81
⇒
4
3
+
64
27
=
−
k
81
⇒
k
=
−
300
© examsnet.com
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