Solve 3x^8-12x^4+1=0

$$3 x ^ { 8 } - 12 x ^ { 4 } + 1 = 0$$

Solve for x (complex solution)

$$x\in \frac{3^{\frac{3}{4}}\sqrt[4]{6-\sqrt{33}}}{3},-\frac{3^{\frac{3}{4}}i\sqrt[4]{6-\sqrt{33}}}{3},\frac{3^{\frac{3}{4}}i\sqrt[4]{6-\sqrt{33}}}{3},-\frac{3^{\frac{3}{4}}\sqrt[4]{6-\sqrt{33}}}{3},-\frac{3^{\frac{3}{4}}i\sqrt[4]{\sqrt{33}+6}}{3},\frac{3^{\frac{3}{4}}\sqrt[4]{\sqrt{33}+6}}{3},\frac{3^{\frac{3}{4}}i\sqrt[4]{\sqrt{33}+6}}{3},-\frac{3^{\frac{3}{4}}\sqrt[4]{\sqrt{33}+6}}{3}$$

Steps Using the Quadratic Formula

Substitute t for x^{4}.

$$3t^{2}-12t+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 3 for a, -12 for b, and 1 for c in the quadratic formula.

$$t=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 3\times 1}}{2\times 3}$$

Do the calculations.

$$t=\frac{12±2\sqrt{33}}{6}$$

Solve the equation t=\frac{12±2\sqrt{33}}{6} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{33}}{3}+2 t=-\frac{\sqrt{33}}{3}+2$$

Since x=t^{4}, the solutions are obtained by solving the equation for each t.

$$x=-i\sqrt[4]{\frac{\sqrt{33}}{3}+2} x=-\sqrt[4]{\frac{\sqrt{33}}{3}+2} x=i\sqrt[4]{\frac{\sqrt{33}}{3}+2} x=\sqrt[4]{\frac{\sqrt{33}}{3}+2} x=-i\sqrt[4]{-\frac{\sqrt{33}}{3}+2} x=-\sqrt[4]{-\frac{\sqrt{33}}{3}+2} x=i\sqrt[4]{-\frac{\sqrt{33}}{3}+2} x=\sqrt[4]{-\frac{\sqrt{33}}{3}+2}$$

Solve for x

$$x=-\frac{3^{\frac{3}{4}}\sqrt[4]{6-\sqrt{33}}}{3}\approx -0.540182845$$

$$x=\frac{3^{\frac{3}{4}}\sqrt[4]{6-\sqrt{33}}}{3}\approx 0.540182845$$

$$x = \frac{3 ^ {\frac{3}{4}} \sqrt[4]{\sqrt{33} + 6}}{3} \approx 1.406626835$$

$$x = -\frac{3 ^ {\frac{3}{4}} \sqrt[4]{\sqrt{33} + 6}}{3} \approx -1.406626835$$

Steps Using the Quadratic Formula

Substitute t for x^{4}.

$$3t^{2}-12t+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 3 for a, -12 for b, and 1 for c in the quadratic formula.

$$t=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 3\times 1}}{2\times 3}$$

Do the calculations.

$$t=\frac{12±2\sqrt{33}}{6}$$

Solve the equation t=\frac{12±2\sqrt{33}}{6} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{33}}{3}+2 t=-\frac{\sqrt{33}}{3}+2$$

Since x=t^{4}, the solutions are obtained by evaluating x=±\sqrt[4]{t} for positive t.

$$x=\frac{\sqrt[4]{27\sqrt{33}+162}}{3} x=-\frac{\sqrt[4]{27\sqrt{33}+162}}{3} x=\frac{\sqrt[4]{162-27\sqrt{33}}}{3} x=-\frac{\sqrt[4]{162-27\sqrt{33}}}{3}$$