Help with finding all real solutions

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To solve the equation x^4 - 8x^2 + 2 = 0, a substitution of t = x^2 simplifies it to a quadratic form: t^2 - 8t + 2 = 0. The quadratic formula can then be applied to find solutions for t, which must be converted back to x by taking the square root of the results. It's crucial to discard any negative roots since x^2 cannot be negative. While complex solutions exist, the focus of the discussion is on finding all real solutions, as indicated by the thread title. Understanding the relationship between complex and real roots is also highlighted as important for tracking solutions.
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Homework Statement


x^4 - 8x^2 + 2 = 0

Homework Equations


quadratic formula?

The Attempt at a Solution


i would've tried using the quadratic formula, but I am not sure if this would work with that seeing as how it's not ax^2 + bx + c
 
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You will need to use the factor/remainder theorem.
 
The example equation is in quadratic form, or something like it. First, you may use a substitution, something like, t=x^2. This can give you like, this:

t^2 - 8t + 2 = 0

Now, you can solve THAT one and you may obtain two solutions, but those solutions are for t. Now, solve each of those solutions for x (remember to first replace t with x^2 )
 
The polynomial fortunately is quadratic, but in x2.

If u = x2 then you have x4 -8x2 + 2 = u2 - 8u + 2 = 0.

You can use the quadratic formula to solve for x2 keeping in mind that any negative roots from the formula must be discarded since x2 can never be negative.

--Elucidus
 
Elucidus said:
...keeping in mind that any negative roots from the formula must be discarded since x2 can never be negative.

--Elucidus

We have this field now called complex numbers that has negitive squares.
i^2=-1 for example
 
lurflurf said:
We have this field now called complex numbers that has negitive squares.
i^2=-1 for example

The thread title is "help with finding all real solutions" so I figured that complex solutions weren't needed.

--Elucidus
 
Elucidus said:
The thread title is "help with finding all real solutions" so I figured that complex solutions weren't needed.

--Elucidus

Good point. I like to keep track of the complex roots to assist in tracking the real roots. The fundamental theorem of a algebra impies each complex root means one less real root to find.
 
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