Solve 4/9x^6-8/6x^3+3/4=0

$$\frac{ 4 }{ 9 } { x }^{ 6 } - \frac{ 8 }{ 6 } { x }^{ 3 } + \frac{ 3 }{ 4 } =0$$

Solve for x (complex solution)

$$x\in \frac{\sqrt[3]{2}\times 3^{\frac{2}{3}}}{2},\frac{\sqrt[3]{2}\times 3^{\frac{2}{3}}e^{\frac{4\pi i}{3}}}{2},\frac{\sqrt[3]{2}\times 3^{\frac{2}{3}}e^{\frac{2\pi i}{3}}}{2},\frac{\sqrt[3]{6}e^{\frac{4\pi i}{3}}}{2},\frac{\sqrt[3]{6}}{2},\frac{\sqrt[3]{6}e^{\frac{2\pi i}{3}}}{2}$$

Steps Using the Quadratic Formula

Reduce the fraction \frac{8}{6} to lowest terms by extracting and canceling out 2.

$$\frac{4}{9}x^{6}-\frac{4}{3}x^{3}+\frac{3}{4}=0$$

Substitute t for x^{3}.

$$\frac{4}{9}t^{2}-\frac{4}{3}t+\frac{3}{4}=0$$

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 \frac{4}{9} for a, -\frac{4}{3} for b, and \frac{3}{4} for c in the quadratic formula.

$$t=\frac{-\left(-\frac{4}{3}\right)±\sqrt{\left(-\frac{4}{3}\right)^{2}-4\times \frac{4}{9}\times \frac{3}{4}}}{\frac{4}{9}\times 2}$$

Do the calculations.

$$t=\frac{\frac{4}{3}±\frac{2}{3}}{\frac{8}{9}}$$

Solve the equation t=\frac{\frac{4}{3}±\frac{2}{3}}{\frac{8}{9}} when ± is plus and when ± is minus.

$$t=\frac{9}{4} t=\frac{3}{4}$$

Since x=t^{3}, the solutions are obtained by solving the equation for each t.

$$x=-\frac{\sqrt[3]{9}\times 4^{\frac{2}{3}}e^{\frac{\pi i}{3}}}{4} x=\frac{\sqrt[3]{9}\times 4^{\frac{2}{3}}ie^{\frac{\pi i}{6}}}{4} x=\frac{\sqrt[3]{9}\times 4^{\frac{2}{3}}}{4} x=-\frac{\sqrt[3]{3}\times 4^{\frac{2}{3}}e^{\frac{\pi i}{3}}}{4} x=\frac{\sqrt[3]{3}\times 4^{\frac{2}{3}}ie^{\frac{\pi i}{6}}}{4} x=\frac{\sqrt[3]{3}\times 4^{\frac{2}{3}}}{4}$$

Solve for x

$$x=\frac{\sqrt[3]{6}}{2}\approx 0.908560296$$

$$x = \frac{\sqrt[3]{2} \cdot 3 ^ {\frac{2}{3}}}{2} \approx 1.310370697$$

Steps Using the Quadratic Formula

Reduce the fraction \frac{8}{6} to lowest terms by extracting and canceling out 2.

$$\frac{4}{9}x^{6}-\frac{4}{3}x^{3}+\frac{3}{4}=0$$

Substitute t for x^{3}.

$$\frac{4}{9}t^{2}-\frac{4}{3}t+\frac{3}{4}=0$$

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 \frac{4}{9} for a, -\frac{4}{3} for b, and \frac{3}{4} for c in the quadratic formula.

$$t=\frac{-\left(-\frac{4}{3}\right)±\sqrt{\left(-\frac{4}{3}\right)^{2}-4\times \frac{4}{9}\times \frac{3}{4}}}{\frac{4}{9}\times 2}$$

Do the calculations.

$$t=\frac{\frac{4}{3}±\frac{2}{3}}{\frac{8}{9}}$$

Solve the equation t=\frac{\frac{4}{3}±\frac{2}{3}}{\frac{8}{9}} when ± is plus and when ± is minus.

$$t=\frac{9}{4} t=\frac{3}{4}$$

Since x=t^{3}, the solutions are obtained by evaluating x=\sqrt[3]{t} for each t.

$$x=\frac{\sqrt[3]{9}\times 4^{\frac{2}{3}}}{4} x=\frac{\sqrt[3]{3}\times 4^{\frac{2}{3}}}{4}$$