# Help with finding all real solutions

1. Sep 24, 2009

### akp

1. The problem statement, all variables and given/known data
x^4 - 8x^2 + 2 = 0

2. Relevant equations

3. The attempt at a solution
i would've tried using the quadratic formula, but im not sure if this would work with that seeing as how it's not ax^2 + bx + c

2. Sep 24, 2009

### fghtffyrdmns

You will need to use the factor/remainder theorem.

3. Sep 24, 2009

### symbolipoint

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 )

4. Sep 24, 2009

### Elucidus

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

5. Sep 24, 2009

### lurflurf

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

6. Sep 25, 2009

### Elucidus

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

--Elucidus

7. Sep 25, 2009

### lurflurf

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.