Solve 2=a^8+28a^4+36=

$$2 = a ^ { 8 } + 28 a ^ { 4 } + 36 =$$

Solve for a

$$a\in \left(\frac{1}{2}+\frac{1}{2}i\right)\sqrt[4]{36\sqrt{2}+56},\left(\frac{1}{2}-\frac{1}{2}i\right)\sqrt[4]{36\sqrt{2}+56},\left(-\frac{1}{2}+\frac{1}{2}i\right)\sqrt[4]{36\sqrt{2}+56},\left(-\frac{1}{2}-\frac{1}{2}i\right)\sqrt[4]{36\sqrt{2}+56},\left(\frac{1}{2}-\frac{1}{2}i\right)\sqrt[4]{56-36\sqrt{2}},\left(\frac{1}{2}+\frac{1}{2}i\right)\sqrt[4]{56-36\sqrt{2}},\left(-\frac{1}{2}+\frac{1}{2}i\right)\sqrt[4]{56-36\sqrt{2}},\left(-\frac{1}{2}-\frac{1}{2}i\right)\sqrt[4]{56-36\sqrt{2}}$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$a^{8}+28a^{4}+36=2$$

Subtract 2 from both sides.

$$a^{8}+28a^{4}+36-2=0$$

Subtract 2 from 36 to get 34.

$$a^{8}+28a^{4}+34=0$$

Substitute t for a^{4}.

$$t^{2}+28t+34=0$$

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, 28 for b, and 34 for c in the quadratic formula.

$$t=\frac{-28±\sqrt{28^{2}-4\times 1\times 34}}{2}$$

Do the calculations.

$$t=\frac{-28±18\sqrt{2}}{2}$$

Solve the equation t=\frac{-28±18\sqrt{2}}{2} when ± is plus and when ± is minus.

$$t=9\sqrt{2}-14 t=-9\sqrt{2}-14$$

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

$$a=\left(\frac{1}{2}-\frac{1}{2}i\right)\sqrt[4]{56-36\sqrt{2}} a=\left(-\frac{1}{2}-\frac{1}{2}i\right)\sqrt[4]{56-36\sqrt{2}} a=\left(-\frac{1}{2}+\frac{1}{2}i\right)\sqrt[4]{56-36\sqrt{2}} a=\left(\frac{1}{2}+\frac{1}{2}i\right)\sqrt[4]{56-36\sqrt{2}} a=\left(\frac{1}{2}-\frac{1}{2}i\right)\sqrt[4]{36\sqrt{2}+56} a=\left(-\frac{1}{2}-\frac{1}{2}i\right)\sqrt[4]{36\sqrt{2}+56} a=\left(-\frac{1}{2}+\frac{1}{2}i\right)\sqrt[4]{36\sqrt{2}+56} a=\left(\frac{1}{2}+\frac{1}{2}i\right)\sqrt[4]{36\sqrt{2}+56}$$