Solve quartic equation $a^4-6a^2b-8ac-3b^2=0$

Question

Please help me to find roots of this quartic equation for a: $$a^4-6a^2b-8ac-3b^2=0$$

Wolfram Alpha gave this result.

But may be there is simple way to get all a?

Answer 1

Rewrite the equation by completing the squares \begin{align} a^4-6a^2b-8ac-3b^2 =(a^2+s)^2 -( 2s+6b)\left( a +\frac{4c}{2s+6b} \right)^2\tag1\\ \end{align} where $s$ happens to satisfy $(s+b)^3=8(c^2-b^3)$, or $s=2\sqrt[3]{c^2-b^3}-b$. For notational convenience, denote $$p=\frac12\sqrt{2s+6b}= \sqrt{\sqrt[3]{c^2-b^3}+b }\tag2$$ and then $s=2p^2-3b$. Substitute them into (1) and then factorize to get $$\left(a^2+2pa+2p^2-3b-\frac{2c}p\right) \left(a^2-2pa+2p^2-3b+\frac{2c}p\right)=0 $$ which yields the solutions $$a=p\pm \sqrt{ 3b-p^2+\frac{2c}p },\>\>\> -p\pm \sqrt{ 3b-p^2-\frac{2c}p } $$ with $p$ given by (2).

Answer 2

$$\text{Given}\quad a^4-6a^2b-8ac-3b^2=0\quad\implies 3b^2+6a^2b -(a^4-8ac)=0$$

The solution for $b$ using the quadratic equation is $$b = \pm\frac{2 \sqrt{a^4 - 2 a c}}{\sqrt{3}} - a^2\quad\text{for}\quad a\in\mathbb{R}\land c\in\mathbb{R}\land b\in\mathbb{R}\iff a^4-2ac\ge0$$ Technically, this defines all values of $a$ but perhaps you seek integers.

Regular algebras solves for $c$ as $$c = \frac{a^4 - 6 a^2 b - 3 b^2}{8 a}$$ The second equation looks easier to work with (though you can try the first if you want) and experimentation in a spreadsheet offers these sample integer triples. \begin{equation} (1,-11,-37)\quad (1,-7,-13)\quad (1,-3,-1)\quad (1,1,-1)\quad (1,5,-13)\quad (1,9,-37)\quad\\ (2,-8,1)\quad (2,-4,4)\quad (2,0,1)\quad (2,4,-8)\quad (2,8,-23)\quad (2,12,-44)\quad\\ (3,-11,13)\quad (3,-7,13)\quad (3,-3,9)\quad (3,1,1)\quad (3,5,-11)\quad (3,9,-27)\quad (3,13,-47)\quad\\ (4,-8,26)\quad (4,0,8)\quad (4,8,-22)\quad\\ (5,5,-5)\quad\\ (6,-8,59)\quad (6,-4,44)\quad (6,0,27)\quad (6,4,8)\quad (6,8,-13)\quad (6,12,-36)\quad\\ (7,-7,77)\quad\\ (8,-8,109)\quad (8,0,64)\quad (8,8,13)\quad \end{equation}

WolframAlpha offer only the roots $a=0,b=0$ or $a\ne0\land c = \frac{a^4 - 6 a^2 b - 3 b^2}{8 a}$ as shown here. This solution for $c$ (as I pointed out above) appears to have infinite integer solutions and, when $a=1$, it appears that these have a symmetry for values of $b$ around zero. Let me know if you have other questions about this.