SOLUTION: I need to find all real solutions to this equation: 4x^-4 -16x^-2+4 = 0 This is how a proceeded: 1/4x^4 - 1/16x^2 +4 = 0; I figured x cannot equal zero, and the LCD =

Question 1101465: I need to find all real solutions to this equation:
4x^-4 -16x^-2+4 = 0
This is how a proceeded:
1/4x^4 - 1/16x^2 +4 = 0; I figured x cannot equal zero, and the LCD = 16x^4
I multiplied both sides by the LCD to clear the denominators:
4 - x^2 + 64x^4 = 0
or
64x^4 -x^2 +4 = 0
I made x^2 = w to get a quadratic equation:
64w^2 -w +4 = 0
I tried factoring this, and I couldn't. Then I figured out that the discriminant is less than zero, which will give me imaginary numbers (b^2 -4ac = 1 -1024 = -1023)
This can't possibly be right, so I must be doing something wrong. Can you help?
Thank you
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
find all real solutions to this equation:
4x^-4 -16x^-2+4 = 0
------
Let w = x^-2
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Substitute to get:
4w^2 - 16w + 4 = 0
---------------(----------
Solve for "w"::
4(w^2 - 4w + 1) = 0
----
Quadratic Formula::
w = [4 +- sqrt(16-4*1*1)]/2
w = [4 +- sqrt(12)]/2
w = 2 +- 2sqrt(3)
w = -1.4641 or w = 5.4641
------
Ans:: x = 0.1830.. or x = -0.6830..
------------
Cheers,
Stan H.
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