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?

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 $x^2 + (y + a) x + A=0$
