Solve 2x^4+40x^2+128=0

$$2 x ^ { 4 } + 40 x ^ { 2 } + 128 = 0$$

Solve for x (complex solution)

$$x=2i$$

$$x=-2i$$

$$x=-4i$$

$$x=4i$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$2t^{2}+40t+128=0$$

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 2 for a, 40 for b, and 128 for c in the quadratic formula.

$$t=\frac{-40±\sqrt{40^{2}-4\times 2\times 128}}{2\times 2}$$

Do the calculations.

$$t=\frac{-40±24}{4}$$

Solve the equation t=\frac{-40±24}{4} when ± is plus and when ± is minus.

$$t=-4 t=-16$$

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

$$x=-2i x=2i x=-4i x=4i$$