Solve 4a^4-a^2+4=0

$$4 a ^ { 4 } - a ^ { 2 } + 4 = 0$$

Solve for a

$$a=e^{\frac{-\arctan(3\sqrt{7})i+2\pi i}{2}}\approx -0.75+0.661437828i$$

$$a=e^{-\frac{\arctan(3\sqrt{7})i}{2}}\approx 0.75-0.661437828i$$

$$a=e^{\frac{\arctan(3\sqrt{7})i+2\pi i}{2}}\approx -0.75-0.661437828i$$

$$a=e^{\frac{\arctan(3\sqrt{7})i}{2}}\approx 0.75+0.661437828i$$

Steps Using the Quadratic Formula

Substitute t for a^{2}.

$$4t^{2}-t+4=0$$

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 4 for a, -1 for b, and 4 for c in the quadratic formula.

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

Do the calculations.

$$t=\frac{1±\sqrt{-63}}{8}$$

Solve the equation t=\frac{1±\sqrt{-63}}{8} when ± is plus and when ± is minus.

$$t=\frac{1+3\sqrt{7}i}{8} t=\frac{-3\sqrt{7}i+1}{8}$$

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

$$a=e^{\frac{\arctan(3\sqrt{7})i+2\pi i}{2}} a=e^{\frac{\arctan(3\sqrt{7})i}{2}} a=e^{-\frac{\arctan(3\sqrt{7})i}{2}} a=e^{\frac{-\arctan(3\sqrt{7})i+2\pi i}{2}}$$