(2x) / (1+x^2)^2 = 2/5

I completely forgot how to solve equations with x^4

can someone tell me how to do this?

I take the denominator to the other side and factored it and got an equation with x to the 4th degree.

Printable View

- Sep 22nd 2007, 08:49 PMxfyzHigher degree equations
(2x) / (1+x^2)^2 = 2/5

I completely forgot how to solve equations with x^4

can someone tell me how to do this?

I take the denominator to the other side and factored it and got an equation with x to the 4th degree. - Sep 22nd 2007, 08:52 PMJhevon
what was the equation you got? show us and we'll take it from there, there is no one way to attack a quartic, we must do it by case

EDIT: ok, this is not a "nice" quadratic, it does not have integer roots. so if you have to solve this by hand i see no other way to do it at the moment other than to use the quartic formula. which is a mess. you can use Cardano's or Galois' method (your professor is cruel) - Sep 22nd 2007, 09:47 PMxfyz
- Sep 22nd 2007, 09:49 PMJhevon
- Sep 23rd 2007, 08:13 AMThePerfectHacker
Let $\displaystyle x=\tan y$ then we have,

$\displaystyle \frac{2\tan y}{(1+\tan^2 y)^2} = \frac{2}{5}$

Thus,

$\displaystyle \frac{2\tan y}{\sec^4 y} = \frac{2}{5}$

Thus,

$\displaystyle 2\tan y\cos ^4 y = \frac{2}{5}$

Which means,

$\displaystyle 2 \sin y \cos y \cos^2 y = \frac{2}{5}$

Identities,

$\displaystyle \sin 2y \cos^2 y = \frac{2}{5}$

More,

$\displaystyle (1 - 2\cos^2 y)(\cos^2 y) = \frac{2}{5}$

Let $\displaystyle z=\cos^2 y$ to get,

$\displaystyle (1-2z)z = \frac{2}{5}$.

The above is quadradic which is solvable.

Then you need to convert $\displaystyle \cos $ into $\displaystyle \tan $.

And then take arctangent to get final answer. - Sep 23rd 2007, 08:21 AMtopsquark
- Sep 23rd 2007, 08:22 AMtopsquark
- Sep 23rd 2007, 08:46 AMThePerfectHacker
And it is not called the Galois method. That links derived the method for 4th degree equations using Galois Theory.

- Sep 23rd 2007, 09:24 AMJhevon
i was speaking loosely after all. i wasn't sure what the method's were called. i just named them by the name i saw on the webpage