A strange equation I

Get ready for a lot of substitutions!:

First of all, we will make the equation $x^{4}+8x+12=0$ in the form of two quadratics: $(x^{2}+Px+Q)(x^{2}-Px+R)=0$ . After expanding, we have: $x^{4}+(Q+R-P^{2})x^{2}+P(R-Q)x+QR=0$ Now, form an equation system by comparing the coefficients: $Q+R-P^{2}=0 \implies Q+R=P^{2}\\ P(R-Q)=8 \implies R-Q=\dfrac{8}{P}\\ QR=12$

Solve for $R$ and $Q$ by adding and substracting, respectively, the first and the second equation: $R=\dfrac{P^{2}+\dfrac{8}{P}}{2}\\ Q=\dfrac{P^{2}-\dfrac{8}{P}}{2}$

And substitute them in the third equation, in order to solve for $P$ : $\left(\dfrac{P^{2}+\dfrac{8}{P}}{2}\right)\left(\dfrac{P^{2}-\dfrac{8}{P}}{2}\right)=12\\ P^{6}-48P^{2}-64=0\\ (P^{2})^{3}-48P^{2}-64=0$ Here we have a bi-cubic equation, so let's solve for $P^{2}$ . Now, here comes a tricky part: consider the identity $(u+v)^{3}-3uv(u+v)-(u^{3}+v^{3})=0$ and make it match with the equation: $P^{2}=u+v\\ -3uv=-48 \implies uv=16\\ -(u^{3}+v^{3})=-64 \implies u^{3}+v^{3}=64$

And another trick: cube both sides of the first equation: $uv=16 \implies (u)^{3}(v)^{3}=4096$

Observe that the equation system we formed matches, by Vieta's formulas, with a second degree equation with roots $u^{3}$ and $v^{3}$ , so let $z_{1}=u^{3}$ and $z_{2}=v^{3}$ : $z^{2}-64z+4096=0$ Use the quadratic formula: $z_{1,2}=32\pm32i\sqrt{3}$ Where $i=\sqrt{-1}$ is the maginary unit.

Substitute back for $u$ and $v$ : $u=\sqrt[3]{32+32i\sqrt{3}}\\ v=\sqrt[3]{32-32i\sqrt{3}}$

Express them in polar form, and consider only the principal value: $u=\sqrt[3]{64e^{\frac{\pi i}{3}}} \implies u=4e^{\frac{\pi i}{9}}\\ v=\sqrt[3]{64e^{-\frac{\pi i}{3}}} \implies v=4e^{-\frac{\pi i}{9}}$

Reverse to the rectangular form and convert from radians to degrees: $u=4(\cos(20°)+i \sin(20°))\\ v=4(\cos(20°)-i \sin(20°))$

Now, $P^{2}=u+v$ , so: $P^{2}=4(\cos(20°)+i \sin(20°))+4(\cos(20°)-i \sin(20°))\\ P^{2}=8\cos(20°)\\ P=2\sqrt{2\cos(20°)}$

Substitute back to obtain $R$ and $Q$ : $R=4\cos(20°)+\sqrt{2\sec(20°)}\\ Q=4\cos(20°)-\sqrt{2\sec(20°)}$

Now, in our initial form $(x^{2}+Px+Q)(x^{2}-Px+R)=0$ , solve each factor using the quadratic formula: $x_{1,2}=\dfrac{-P\pm\sqrt{P^{2}-4Q}}{2}\\ x_{3,4}=\dfrac{P\pm\sqrt{P^{2}-4R}}{2}$

And substitute back again with the known values of $P$ , $Q$ and $R$ : $x_{1,2}=-\sqrt{2\cos(20°)}\pm\sqrt{\sqrt{2\sec(20°)}-2\cos(20°)}\\ x_{3,4}=\sqrt{2\cos(20°)}\pm\sqrt{-\sqrt{2\sec(20°)}-2\cos(20°)}$

The four solutions only differ from the signs, so we can conclude: $x_{1,2,3,4}=\mp\sqrt{2\cos(20°)}\pm\sqrt{\pm\sqrt{2\sec(20°)}-2\cos(20°)}$ With the correct sign choose for each $x$ : $x_{1}=-++\\ x_{2}=--+\\ x_{3}=++-\\ x_{4}=+--$

So, after all, $a=b=c=2$ , $d=20$ , and $a+b+c+d=\boxed{26}$ .

Revised Ferrari's Method:

x^4 + 8 x + 12 = 0

=> x^4 + p x^2 - p x^2 + 8 x + 12 = 0

=> (x^2 + p/ 2)^2 - (p x^2 - 8 x) + 12 - p^2/ 4 = 0

=> (x^2 + p/ 2)^2 - [sqrt(p) x - 4/ sqrt(p)]^2 + 12 - p^2/ 4 - 16/ p = 0

With 12 - p^2/ 4 - 16/ p = 0 such that p^3 - 48 p - 64 = 0 where p = 8 Cos 20 deg is taken,

(x^2 + p/ 2)^2 - [sqrt(p) x - 4/ sqrt(p)]^2 = 0

=> [x^2 + p/ 2 - sqrt(p) x + 4/ sqrt(p)][x^2 + p/ 2 + sqrt(p) x - 4/ sqrt(p)] = 0

=> [x^2 - sqrt(p) x + 4/ sqrt(p) + p/ 2][x^2 + sqrt(p) x - 4/ sqrt(p) + p/ 2 ] = 0

=> [(x - sqrt(p)/ 2)^2 - (p/ 4 - 4/ sqrt(p) - p/ 2)][(x + sqrt(p)/ 2)^2 - (p/ 4 + 4/ sqrt(p) - p/ 2)] = 0

=> [(x - sqrt(p)/ 2)^2 - (- 4/ sqrt(p) - p/ 4)][(x + sqrt(p)/ 2)^2 - (4/ sqrt(p) - p/ 4)] = 0

Hence,

x = sqrt(p)/ 2 - sqrt(- 4/ sqrt(p) - p/ 4) or sqrt(p)/ 2 + sqrt(- 4/ sqrt(p) - p/ 4) OR - sqrt(p)/ 2 - sqrt(4/ sqrt(p) - p/ 4) or - sqrt(p)/ 2 + sqrt(4/ sqrt(p) - p/ 4)

With p = 8 Cos (20 deg) as 0 deg < d deg < 40 deg, {a degree is equal to Pi/180 and therefore never repeat wrongly. The question made a slight mistake.}

x = Sign sqrt(2 Cos 20 deg) +/- sqrt[-Sign sqrt(2 Sec 20 deg) - 2 Cos 20 deg]

a + b + c + d = 2 + 2 + 2 + 20 = 26.