Solve 5x^4+3x^2+1=-1

$$5 x ^ { 4 } + 3 x ^ { 2 } + 1 = - 1$$

Solve for x (complex solution)

$$x=\frac{\sqrt[4]{250}e^{\frac{\left(\arctan(\frac{\sqrt{31}}{3})+\pi \right)i}{2}}}{5}\approx -0.407710395+0.682808733i$$

$$x=\frac{\sqrt[4]{250}e^{\frac{\arctan(\frac{\sqrt{31}}{3})i+3\pi i}{2}}}{5}\approx 0.407710395-0.682808733i$$

$$x=\frac{\sqrt[4]{250}e^{\frac{-\arctan(\frac{\sqrt{31}}{3})i+3\pi i}{2}}}{5}\approx -0.407710395-0.682808733i$$

$$x=\frac{\sqrt[4]{250}e^{\frac{-\arctan(\frac{\sqrt{31}}{3})i+\pi i}{2}}}{5}\approx 0.407710395+0.682808733i$$

Steps Using the Quadratic Formula

Add 1 to both sides.

$$5x^{4}+3x^{2}+1+1=0$$

Add 1 and 1 to get 2.

$$5x^{4}+3x^{2}+2=0$$

Substitute t for x^{2}.

$$5t^{2}+3t+2=0$$

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 5 for a, 3 for b, and 2 for c in the quadratic formula.

$$t=\frac{-3±\sqrt{3^{2}-4\times 5\times 2}}{2\times 5}$$

Do the calculations.

$$t=\frac{-3±\sqrt{-31}}{10}$$

Solve the equation t=\frac{-3±\sqrt{-31}}{10} when ± is plus and when ± is minus.

$$t=\frac{-3+\sqrt{31}i}{10} t=\frac{-\sqrt{31}i-3}{10}$$

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

$$x=\frac{\sqrt[4]{250}e^{\frac{-\arctan(\frac{\sqrt{31}}{3})i+3\pi i}{2}}}{5} x=\frac{\sqrt[4]{250}e^{\frac{-\arctan(\frac{\sqrt{31}}{3})i+\pi i}{2}}}{5} x=\frac{\sqrt[4]{250}e^{\frac{\arctan(\frac{\sqrt{31}}{3})i+3\pi i}{2}}}{5} x=\frac{\sqrt[4]{250}e^{\frac{\left(\arctan(\frac{\sqrt{31}}{3})+\pi \right)i}{2}}}{5}$$