Solve 11k^4-9k^2+2=0

$$11 k ^ { 4 } - 9 k ^ { 2 } + 2 = 0$$

Solve for k

$$k=\frac{\sqrt[4]{2662}e^{\frac{-\arctan(\frac{\sqrt{7}}{9})i+2\pi i}{2}}}{11}\approx -0.646332864+0.09303366i$$

$$k=\frac{\sqrt[4]{2662}e^{-\frac{\arctan(\frac{\sqrt{7}}{9})i}{2}}}{11}\approx 0.646332864-0.09303366i$$

$$k=\frac{\sqrt[4]{2662}e^{\frac{\arctan(\frac{\sqrt{7}}{9})i+2\pi i}{2}}}{11}\approx -0.646332864-0.09303366i$$

$$k=\frac{\sqrt[4]{2662}e^{\frac{\arctan(\frac{\sqrt{7}}{9})i}{2}}}{11}\approx 0.646332864+0.09303366i$$

Steps Using the Quadratic Formula

Substitute t for k^{2}.

$$11t^{2}-9t+2=0$$

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 11 for a, -9 for b, and 2 for c in the quadratic formula.

$$t=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 11\times 2}}{2\times 11}$$

Do the calculations.

$$t=\frac{9±\sqrt{-7}}{22}$$

Solve the equation t=\frac{9±\sqrt{-7}}{22} when ± is plus and when ± is minus.

$$t=\frac{9+\sqrt{7}i}{22} t=\frac{-\sqrt{7}i+9}{22}$$

Since k=t^{2}, the solutions are obtained by evaluating k=±\sqrt{t} for each t.

$$k=\frac{\sqrt[4]{2662}e^{\frac{\arctan(\frac{\sqrt{7}}{9})i+2\pi i}{2}}}{11} k=\frac{\sqrt[4]{2662}e^{\frac{\arctan(\frac{\sqrt{7}}{9})i}{2}}}{11} k=\frac{\sqrt[4]{2662}e^{-\frac{\arctan(\frac{\sqrt{7}}{9})i}{2}}}{11} k=\frac{\sqrt[4]{2662}e^{\frac{-\arctan(\frac{\sqrt{7}}{9})i+2\pi i}{2}}}{11}$$