Solve 2x^4+8x^2+7=0

$$2 x ^ { 4 } + 8 x ^ { 2 } + 7 = 0$$

Solve for x (complex solution)

$$x=\frac{i\sqrt{2\sqrt{2}+8}}{2}\approx 1.645328776i$$

$$x=-\frac{i\sqrt{2\sqrt{2}+8}}{2}\approx -0-1.645328776i$$

$$x=-\frac{i\sqrt{8-2\sqrt{2}}}{2}\approx -0-1.137054624i$$

$$x=\frac{i\sqrt{8-2\sqrt{2}}}{2}\approx 1.137054624i$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$2t^{2}+8t+7=0$$

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, 8 for b, and 7 for c in the quadratic formula.

$$t=\frac{-8±\sqrt{8^{2}-4\times 2\times 7}}{2\times 2}$$

Do the calculations.

$$t=\frac{-8±2\sqrt{2}}{4}$$

Solve the equation t=\frac{-8±2\sqrt{2}}{4} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{2}}{2}-2 t=-\frac{\sqrt{2}}{2}-2$$

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

$$x=-i\sqrt{-\frac{\sqrt{2}}{2}+2} x=i\sqrt{-\frac{\sqrt{2}}{2}+2} x=-i\sqrt{\frac{\sqrt{2}}{2}+2} x=i\sqrt{\frac{\sqrt{2}}{2}+2}$$