Solve x^4-3x^2-9=0

$$x ^ { 4 } - 3 x ^ { 2 } - 9 = 0$$

Solve for x (complex solution)

$$x=\frac{i\sqrt{6\left(\sqrt{5}-1\right)}}{2}\approx 1.361654129i$$

$$x=-\frac{i\sqrt{6\left(\sqrt{5}-1\right)}}{2}\approx -0-1.361654129i$$

$$x = -\frac{\sqrt{6 {(\sqrt{5} + 1)}}}{2} \approx -2.203202661$$

$$x = \frac{\sqrt{6 {(\sqrt{5} + 1)}}}{2} \approx 2.203202661$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$t^{2}-3t-9=0$$

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, -3 for b, and -9 for c in the quadratic formula.

$$t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\left(-9\right)}}{2}$$

Do the calculations.

$$t=\frac{3±3\sqrt{5}}{2}$$

Solve the equation t=\frac{3±3\sqrt{5}}{2} when ± is plus and when ± is minus.

$$t=\frac{3\sqrt{5}+3}{2} t=\frac{3-3\sqrt{5}}{2}$$

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

$$x=-\sqrt{\frac{3\sqrt{5}+3}{2}} x=\sqrt{\frac{3\sqrt{5}+3}{2}} x=-i\sqrt{-\frac{3-3\sqrt{5}}{2}} x=i\sqrt{-\frac{3-3\sqrt{5}}{2}}$$

Solve for x

$$x = -\frac{\sqrt{6 {(\sqrt{5} + 1)}}}{2} \approx -2.203202661$$

$$x = \frac{\sqrt{6 {(\sqrt{5} + 1)}}}{2} \approx 2.203202661$$

Steps Using the Quadratic Formula

Substitute t for x^{2}.

$$t^{2}-3t-9=0$$

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, -3 for b, and -9 for c in the quadratic formula.

$$t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\left(-9\right)}}{2}$$

Do the calculations.

$$t=\frac{3±3\sqrt{5}}{2}$$

Solve the equation t=\frac{3±3\sqrt{5}}{2} when ± is plus and when ± is minus.

$$t=\frac{3\sqrt{5}+3}{2} t=\frac{3-3\sqrt{5}}{2}$$

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

$$x=\frac{\sqrt{6\sqrt{5}+6}}{2} x=-\frac{\sqrt{6\sqrt{5}+6}}{2}$$