The polynomial I'm looking to solve is
$a^4 - \alpha^4 a^2 -k^4 = 0$
The four roots of this polynomial are given as
$a=\pm \lambda_1$ (pair of real roots)
$a=\pm \lambda_2$ (pair of imaginary roots)
where
$\lambda_1=\sqrt{\frac{\alpha^2}{2}+\sqrt{k^4+\frac{\alpha^4}{4}}}$
$\lambda_2=\sqrt{-\frac{\alpha^2}{2}+\sqrt{k^4+\frac{\alpha^4}{4}}}$
To obtain the above solution, I substituted $a^2=D$ and this makes the equation quadratic. I obtained $D=\frac{\alpha^2}{2} \pm \sqrt{k^4+\frac{\alpha^4}{4}}$. No problem till here. But obtaining the solution for $a$ (as above) is confusing to me.
The four roots given above are correct as I checked with some substitution using the expressions obtained from MATLAB and the above. But I'm interested to know how to obtain it.
Any help on the step-by-step procedure from $D$ to $a$ will be highly appreciated.
Thank you very much.
The biquadratic or quartic equation given in your problem statement is still wrong. The $\lambda$ values given are the four roots of $$a^4-\alpha^2 a^2-k^4=0.$$
Assuming that the error is a typo, we continue as shown below. You have obtain the two values of $D$ that satisfy the quadratic equation, $$D^2-\alpha^2 D-k^4=0.$$
They are, as you have mentioned, $$D_1=\dfrac{\alpha^2}{2}+\sqrt{k^4+\dfrac{\alpha^4}{4}}\space\space\qquad\text{and}$$ $$D_2=\dfrac{\alpha^2}{2}-\sqrt{k^4+\dfrac{\alpha^4}{4}}.\qquad\qquad$$
Not only the value of $k^4+\dfrac{\alpha^4}{4}$ is positive for all values of $\alpha$ and $k$, but also $ \dfrac{\alpha^4}{4} \le k^4+\dfrac{\alpha^4}{4}$. In other words, $$\dfrac{\alpha^2}{4} \le \sqrt{k^4+\dfrac{\alpha^4}{4}}.$$
Because of this the value of $D_2$ is negative for all values of $\alpha$ and $k$.
We use $D_1$ to obtain the pair of real roots as, $$a =\pm\sqrt{D_1}=\pm\sqrt{\dfrac{\alpha^2}{2}+\sqrt{k^4+\dfrac{\alpha^4}{4}}}.$$
They agree with the values given as $\lambda_1$.
When it comes to obtaining roots of $a$ using $D_2$, we have a bother, because $D_2$ is negative. Therefore, we express $D_2$ as $$D_2=\left(-1\right)\left(-\dfrac{\alpha^2}{2}+\sqrt{k^4+\dfrac{\alpha^4}{4}}\right).$$
Now, we shall write, $$a =\pm\sqrt{D_2}=\pm\sqrt{\left(-1\right)}\left(\sqrt{-\dfrac{\alpha^2}{2} +\sqrt{k^4+\dfrac{\alpha^4}{4}}}\right).$$
Here, $\sqrt{\left(-1\right)}$ indicates that these two roots are imaginary.