Solve a^4-128a^2+4420=0

$$a ^ { 4 } - 128 a ^ { 2 } + 4420 = 0$$

Solve for a

$$a=\sqrt{2}\sqrt[4]{1105}e^{\frac{-\arctan(\frac{9}{32})i+2\pi i}{2}}\approx -8.07722355+1.114244263i$$

$$a=\sqrt{2}\sqrt[4]{1105}e^{-\frac{\arctan(\frac{9}{32})i}{2}}\approx 8.07722355-1.114244263i$$

$$a=\sqrt{2}\sqrt[4]{1105}e^{\frac{\arctan(\frac{9}{32})i+2\pi i}{2}}\approx -8.07722355-1.114244263i$$

$$a=\sqrt{2}\sqrt[4]{1105}e^{\frac{\arctan(\frac{9}{32})i}{2}}\approx 8.07722355+1.114244263i$$

Steps Using the Quadratic Formula

Substitute t for a^{2}.

$$t^{2}-128t+4420=0$$

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 1 for a, -128 for b, and 4420 for c in the quadratic formula.

$$t=\frac{-\left(-128\right)±\sqrt{\left(-128\right)^{2}-4\times 1\times 4420}}{2}$$

Do the calculations.

$$t=\frac{128±\sqrt{-1296}}{2}$$

Solve the equation t=\frac{128±\sqrt{-1296}}{2} when ± is plus and when ± is minus.

$$t=64+18i t=64-18i$$

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

$$a=\sqrt{2}\sqrt[4]{1105}e^{\frac{\arctan(\frac{9}{32})i+2\pi i}{2}} a=\sqrt{2}\sqrt[4]{1105}e^{\frac{\arctan(\frac{9}{32})i}{2}} a=\sqrt{2}\sqrt[4]{1105}e^{-\frac{\arctan(\frac{9}{32})i}{2}} a=\sqrt{2}\sqrt[4]{1105}e^{\frac{-\arctan(\frac{9}{32})i+2\pi i}{2}}$$