1. 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.

2. 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)

3. 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

4. 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

5. 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.

6. 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

7. Originally Posted by Jhevon
...you can use Cardano's [method]...
For the record, Cardano's method solves cubic equations, not quartic.

-Dan

8. And it is not called the Galois method. That links derived the method for 4th degree equations using Galois Theory.

9. 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