How to solve this quartic? I have the following equation, which I've put into the form of a normal quartic:
$$x^4+4ax^3+(4a^2+1)x^2-1=0$$
I'm trying to find the solutions for $x$ in terms of $a$. Is there an easier way then using the quartic formula?
 A: Solving by Resultant.
$Res_x(x^4 + 4 a x^3 + (4 a^2 + 1) x^2 - 1, x^2 + (y + a) x + A)=\\
1 - a^2 - a^4 + 2 A - A^2 + 5 a^2 A^2 + 4 a^4 A^2 - 2 A^3 + 4 a^2 A^3 + A^4\\
-2 a (1 + 2 A + A^2 + 4 a^2 A^2 + 2 A^3)y + (-1 + 2 a^2 + 4 A + A^2 + 4 a^2 A^2) y^2 - y^4$
Let linear term equal zero, then:
1) find $A$ from cubic $1 + 2 A + A^2 + 4 a^2 A^2 + 2 A^3=0$
2) find $y$ from biquadratic
$1 - a^2 - a^4 + 2 A - A^2 + 5 a^2 A^2+ 4 a^4 A^2 - 2 A^3 + 4 a^2 A^3 + A^4\\
+ (-1 + 2 a^2 + 4 A + A^2 + 4 a^2 A^2) y^2 - y^4=0$
3) find $x$ from quadratic $x^2 + (y + a) x + A=0$
A: Replace $x$ by $y-a$ to get rid of the cube term. Check if the linear term doesn't appear. If yes, then you can put $y^2=t$ and reduce it to a quadratic equation in $t.$
A: It is quadratic in $a$ and so it will be easier to solve in terms of $a$.
$$4x^2a^2+4x^3a+x^4+x^2-1=0$$
Applying the quadratic formula then gives:
$$a=\frac{-x^2\pm\sqrt{1-x^2}}{2x}$$
As the discriminant is not a perfect square, I doubt $x$ can be solved for any more nicely than the result given by the quartic formula unless $a=0$.
A: Not an answer, but maybe useful.
If $\,a=-\frac12\,$ then the equation has the root $\,x=1$.
If $\,a=+\frac12\,$ then the equation has the root $\,x=-1$.
