Solve 3x^8-18x^4+1=0

$$3 x ^ { 8 } - 18 x ^ { 4 } + 1 = 0$$

Solve for x (complex solution)

$$x\in \frac{3^{\frac{3}{4}}\sqrt[4]{9-\sqrt{78}}}{3},-\frac{3^{\frac{3}{4}}i\sqrt[4]{9-\sqrt{78}}}{3},\frac{3^{\frac{3}{4}}i\sqrt[4]{9-\sqrt{78}}}{3},-\frac{3^{\frac{3}{4}}\sqrt[4]{9-\sqrt{78}}}{3},-\frac{3^{\frac{3}{4}}i\sqrt[4]{\sqrt{78}+9}}{3},\frac{3^{\frac{3}{4}}\sqrt[4]{\sqrt{78}+9}}{3},\frac{3^{\frac{3}{4}}i\sqrt[4]{\sqrt{78}+9}}{3},-\frac{3^{\frac{3}{4}}\sqrt[4]{\sqrt{78}+9}}{3}$$

Steps Using the Quadratic Formula

Substitute t for x^{4}.

$$3t^{2}-18t+1=0$$

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, -18 for b, and 1 for c in the quadratic formula.

$$t=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 3\times 1}}{2\times 3}$$

Do the calculations.

$$t=\frac{18±2\sqrt{78}}{6}$$

Solve the equation t=\frac{18±2\sqrt{78}}{6} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{78}}{3}+3 t=-\frac{\sqrt{78}}{3}+3$$

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

$$x=-i\sqrt[4]{\frac{\sqrt{78}}{3}+3} x=-\sqrt[4]{\frac{\sqrt{78}}{3}+3} x=i\sqrt[4]{\frac{\sqrt{78}}{3}+3} x=\sqrt[4]{\frac{\sqrt{78}}{3}+3} x=-i\sqrt[4]{-\frac{\sqrt{78}}{3}+3} x=-\sqrt[4]{-\frac{\sqrt{78}}{3}+3} x=i\sqrt[4]{-\frac{\sqrt{78}}{3}+3} x=\sqrt[4]{-\frac{\sqrt{78}}{3}+3}$$

Solve for x

$$x=-\frac{\sqrt[4]{243-27\sqrt{78}}}{3}\approx -0.486632872$$

$$x=\frac{\sqrt[4]{243-27\sqrt{78}}}{3}\approx 0.486632872$$

$$x = \frac{\sqrt[4]{27 \sqrt{78} + 243}}{3} \approx 1.56141463$$

$$x = -\frac{\sqrt[4]{27 \sqrt{78} + 243}}{3} \approx -1.56141463$$

Steps Using the Quadratic Formula

Substitute t for x^{4}.

$$3t^{2}-18t+1=0$$

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, -18 for b, and 1 for c in the quadratic formula.

$$t=\frac{-\left(-18\right)±\sqrt{\left(-18\right)^{2}-4\times 3\times 1}}{2\times 3}$$

Do the calculations.

$$t=\frac{18±2\sqrt{78}}{6}$$

Solve the equation t=\frac{18±2\sqrt{78}}{6} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{78}}{3}+3 t=-\frac{\sqrt{78}}{3}+3$$

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

$$x=\frac{\sqrt[4]{27\sqrt{78}+243}}{3} x=-\frac{\sqrt[4]{27\sqrt{78}+243}}{3} x=\frac{\sqrt[4]{243-27\sqrt{78}}}{3} x=-\frac{\sqrt[4]{243-27\sqrt{78}}}{3}$$