# Higher degree equations

• Sep 22nd 2007, 09:49 PM
xfyz
Higher 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, 09:52 PM
Jhevon
Quote:

Originally Posted by xfyz
(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.

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, 10:47 PM
xfyz
Quote:

Originally Posted by Jhevon
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)

I have to solve for x for the equation above.
I did it on the calculator but i thought there was a way to do it by hand.
I guess not! (I don't think my teacher wants me to use the quartic..) lol, I think calculator is fine.. I hope
• Sep 22nd 2007, 10:49 PM
Jhevon
Quote:

Originally Posted by xfyz
I have to solve for x for the equation above.
I did it on the calculator but i thought there was a way to do it by hand.
I guess not! (I don't think my teacher wants me to use the quartic..) lol, I think calculator is fine.. I hope

ok, good. no worries then. because doing it by hand, as you can see, is a pain. sometimes we can do it pretty easily by hand, but not in this case
• Sep 23rd 2007, 09:13 AM
ThePerfectHacker
Quote:

Originally Posted by xfyz
(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.

Let $x=\tan y$ then we have,
$\frac{2\tan y}{(1+\tan^2 y)^2} = \frac{2}{5}$
Thus,
$\frac{2\tan y}{\sec^4 y} = \frac{2}{5}$
Thus,
$2\tan y\cos ^4 y = \frac{2}{5}$
Which means,
$2 \sin y \cos y \cos^2 y = \frac{2}{5}$
Identities,
$\sin 2y \cos^2 y = \frac{2}{5}$
More,
$(1 - 2\cos^2 y)(\cos^2 y) = \frac{2}{5}$
Let $z=\cos^2 y$ to get,
$(1-2z)z = \frac{2}{5}$.
Then you need to convert $\cos$ into $\tan$.
And then take arctangent to get final answer.
• Sep 23rd 2007, 09:21 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
Let $x=\tan y$ then we have,
$\frac{2\tan y}{(1+\tan^2 y)^2} = \frac{2}{5}$
Thus,
$\frac{2\tan y}{\sec^4 y} = \frac{2}{5}$
Thus,
$2\tan y\cos ^4 y = \frac{2}{5}$
Which means,
$2 \sin y \cos y \cos^2 y = \frac{2}{5}$
Identities,
$\sin 2y \cos^2 y = \frac{2}{5}$
More,
$(1 - 2\cos^2 y)(\cos^2 y) = \frac{2}{5}$
Let $z=\cos^2 y$ to get,
$(1-2z)z = \frac{2}{5}$.
Then you need to convert $\cos$ into $\tan$.
And then take arctangent to get final answer.

Clever! I always forget to look for something like this.

-Dan
• Sep 23rd 2007, 09:22 AM
topsquark
Quote:

Originally Posted by Jhevon
...you can use Cardano's [method]...

For the record, Cardano's method solves cubic equations, not quartic.

-Dan
• Sep 23rd 2007, 09:46 AM
ThePerfectHacker
And it is not called the Galois method. That links derived the method for 4th degree equations using Galois Theory.
• Sep 23rd 2007, 10:24 AM
Jhevon
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