Solve 20x^4+x^2-2=0

$$20 x ^ { 4 } + x ^ { 2 } - 2 = 0$$

Solve for x (complex solution)

$$x=\frac{i\sqrt{10\left(\sqrt{161}+1\right)}}{20}\approx 0.584990973i$$

$$x=-\frac{i\sqrt{10\left(\sqrt{161}+1\right)}}{20}\approx -0-0.584990973i$$

$$x=-\frac{\sqrt{10\left(\sqrt{161}-1\right)}}{20}\approx -0.540568625$$

$$x=\frac{\sqrt{10\left(\sqrt{161}-1\right)}}{20}\approx 0.540568625$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$20t^{2}+t-2=0$$

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 20 for a, 1 for b, and -2 for c in the quadratic formula.

$$t=\frac{-1±\sqrt{1^{2}-4\times 20\left(-2\right)}}{2\times 20}$$

Do the calculations.

$$t=\frac{-1±\sqrt{161}}{40}$$

Solve the equation t=\frac{-1±\sqrt{161}}{40} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{161}-1}{40} t=\frac{-\sqrt{161}-1}{40}$$

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

$$x=-\sqrt{\frac{\sqrt{161}-1}{40}} x=\sqrt{\frac{\sqrt{161}-1}{40}} x=-i\sqrt{\frac{\sqrt{161}+1}{40}} x=i\sqrt{\frac{\sqrt{161}+1}{40}}$$

Solve for x

$$x=-\frac{\sqrt{10\left(\sqrt{161}-1\right)}}{20}\approx -0.540568625$$

$$x=\frac{\sqrt{10\left(\sqrt{161}-1\right)}}{20}\approx 0.540568625$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$20t^{2}+t-2=0$$

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 20 for a, 1 for b, and -2 for c in the quadratic formula.

$$t=\frac{-1±\sqrt{1^{2}-4\times 20\left(-2\right)}}{2\times 20}$$

Do the calculations.

$$t=\frac{-1±\sqrt{161}}{40}$$

Solve the equation t=\frac{-1±\sqrt{161}}{40} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{161}-1}{40} t=\frac{-\sqrt{161}-1}{40}$$

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

$$x=\frac{\sqrt{\frac{\sqrt{161}-1}{10}}}{2} x=-\frac{\sqrt{\frac{\sqrt{161}-1}{10}}}{2}$$