(x-2/x+2)^2-4(x-2/x+2)+3=0 Solve for x.
Answers
Answered by
12
Explanation:
First making
y
=
x
2
+
2
x
and substituting
y
2
−
2
y
−
3
=
0
Solving for
y
gives
y
=
{
−
1
,
3
}
The next step is to solve
1)
x
2
+
2
x
=
−
1
giving two equal roots
x
=
{
−
1
,
−
1
}
2)
x
2
+
2
x
=
3
giving
x
=
{
−
3
,
1
}
The solutions are
x
=
{
−
3
,
−
1
,
1
}
also we have
(
x
2
+
2
x
)
2
−
2
(
x
2
+
2
x
)
−
3
=
(
x
+
3
)
(
x
+
1
)
2
(
x
−
1
Answered by
1
Answer:
(2^x)^2 + (2^x )(2^2) - 32 = 0
let y = 2^x, replace the value in above eq.
y^2 + 4y - 32 = 0
y^2 - 4y + 8y -32 = 0
y ( y - 4 ) + 8(y -4 ) = 0
(y - 4)(y + 8) =0
hence
y = 4 or -8
but -8 value leads to imaginary number, hence considering only value 4.
y = 4
replace y with 2^x
2^x = 4
2^x. = 2^2
x = 2
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