Thread: x^4-x^2+2=0 Solve. Need help :(

1. x^4-4x^2+2=0

Hey all, the problem in the title is what has me confused. i know there are solutions, but I do not know how to arrive at the solution. Maybe a step by step, or even a website explaining how to solve this type of problem would be very helpful. Thanks : )

Thanks for helping me icemanfan, I really appreciate the time you took to help me. I made a mistake in my original equation, and for that I am truly sorry. The correct equation is x^4-4x^2+2=0. Again, I'm sorry

2. $x^4 - x^2 + 2 = 0$

Substitute $u = x^2$. Then the equation becomes:

$u^2 - u + 2 = 0$

$u = \frac{1 \pm \sqrt{(-1)^2 - 4(1)(2)}}{2}$

$u = \frac{1 \pm \sqrt{-7}}{2}$

So u is a complex number.

Since $u = x^2$ has no real solutions in x, the original equation has no real solutions in x.

3. Still follow icemanfan suggestion but just change it too

$u^2 - 4u + 2~=~0$

$u = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(2)}}{2}$