Solve 5x^4+4x^2-1=0

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

Solve for x (complex solution)

$$x=\frac{\sqrt{5}}{5}\approx 0.447213595$$

$$x=-\frac{\sqrt{5}}{5}\approx -0.447213595$$

$$x=-i$$

$$x=i$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$5t^{2}+4t-1=0$$

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, 4 for b, and -1 for c in the quadratic formula.

$$t=\frac{-4±\sqrt{4^{2}-4\times 5\left(-1\right)}}{2\times 5}$$

Do the calculations.

$$t=\frac{-4±6}{10}$$

Solve the equation t=\frac{-4±6}{10} when ± is plus and when ± is minus.

$$t=\frac{1}{5} t=-1$$

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

$$x=-\frac{\sqrt{5}}{5} x=\frac{\sqrt{5}}{5} x=-i x=i$$

Solve for x

$$x=-\frac{\sqrt{5}}{5}\approx -0.447213595$$

$$x=\frac{\sqrt{5}}{5}\approx 0.447213595$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$5t^{2}+4t-1=0$$

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, 4 for b, and -1 for c in the quadratic formula.

$$t=\frac{-4±\sqrt{4^{2}-4\times 5\left(-1\right)}}{2\times 5}$$

Do the calculations.

$$t=\frac{-4±6}{10}$$

Solve the equation t=\frac{-4±6}{10} when ± is plus and when ± is minus.

$$t=\frac{1}{5} t=-1$$

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

$$x=\frac{\sqrt{5}}{5} x=-\frac{\sqrt{5}}{5}$$