Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
The imaginary roots of the equation (x2+2)2+8 x2=6 x(x2+2) are
Q. The imaginary roots of the equation
(
x
2
+
2
)
2
+
8
x
2
=
6
x
(
x
2
+
2
)
are
546
167
Complex Numbers and Quadratic Equations
Report Error
A
1
±
i
B
2
±
1
C
−
1
±
i
D
−
2
±
i
Solution:
(
x
2
+
2
)
2
+
8
x
2
=
6
x
(
x
2
+
2
)
(
x
2
+
2
)
2
−
6
x
(
x
2
+
2
)
+
8
x
2
=
0
x
2
+
2
=
2
6
x
±
36
x
2
−
32
x
2
x
2
+
2
==
3
x
±
x
x
2
+
2
=
4
x
,
D
>
0
roots are real
x
2
+
2
=
2
x
x
=
1
±
i