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

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

Solve for x (complex solution)

$$x=\frac{i\sqrt{2\left(\sqrt{41}-5\right)}}{2}\approx 0.837593051i$$

$$x=-\frac{i\sqrt{2\left(\sqrt{41}-5\right)}}{2}\approx -0-0.837593051i$$

$$x = -\frac{\sqrt{2 {(\sqrt{41} + 5)}}}{2} \approx -2.387794405$$

$$x = \frac{\sqrt{2 {(\sqrt{41} + 5)}}}{2} \approx 2.387794405$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

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

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 1 for a, -5 for b, and -4 for c in the quadratic formula.

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

Do the calculations.

$$t=\frac{5±\sqrt{41}}{2}$$

Solve the equation t=\frac{5±\sqrt{41}}{2} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{41}+5}{2} t=\frac{5-\sqrt{41}}{2}$$

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

$$x=-\sqrt{\frac{\sqrt{41}+5}{2}} x=\sqrt{\frac{\sqrt{41}+5}{2}} x=-i\sqrt{-\frac{5-\sqrt{41}}{2}} x=i\sqrt{-\frac{5-\sqrt{41}}{2}}$$

Solve for x

$$x = -\frac{\sqrt{2 {(\sqrt{41} + 5)}}}{2} \approx -2.387794405$$

$$x = \frac{\sqrt{2 {(\sqrt{41} + 5)}}}{2} \approx 2.387794405$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

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

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 1 for a, -5 for b, and -4 for c in the quadratic formula.

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

Do the calculations.

$$t=\frac{5±\sqrt{41}}{2}$$

Solve the equation t=\frac{5±\sqrt{41}}{2} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{41}+5}{2} t=\frac{5-\sqrt{41}}{2}$$

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

$$x=\frac{\sqrt{2\sqrt{41}+10}}{2} x=-\frac{\sqrt{2\sqrt{41}+10}}{2}$$