Solve x^4-x^2=2

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

Solve for x (complex solution)

$$x=\sqrt{2}\approx 1.414213562$$

$$x=-\sqrt{2}\approx -1.414213562$$

$$x=-i$$

$$x=i$$

Steps Using the Quadratic Formula

Subtract 2 from both sides.

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

Substitute t for x^{2}.

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

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

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

Do the calculations.

$$t=\frac{1±3}{2}$$

Solve the equation t=\frac{1±3}{2} when ± is plus and when ± is minus.

$$t=2 t=-1$$

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

$$x=-\sqrt{2} x=\sqrt{2} x=-i x=i$$

Solve for x

$$x=-\sqrt{2}\approx -1.414213562$$

$$x=\sqrt{2}\approx 1.414213562$$

Steps Using the Quadratic Formula

Subtract 2 from both sides.

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

Substitute t for x^{2}.

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

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

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

Do the calculations.

$$t=\frac{1±3}{2}$$

Solve the equation t=\frac{1±3}{2} when ± is plus and when ± is minus.

$$t=2 t=-1$$

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

$$x=\sqrt{2} x=-\sqrt{2}$$