Solve x^4-2x^2=3

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

Solve for x (complex solution)

$$x=\sqrt{3}\approx 1.732050808$$

$$x=-\sqrt{3}\approx -1.732050808$$

$$x=-i$$

$$x=i$$

Steps Using the Quadratic Formula

Subtract 3 from both sides.

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

Substitute t for x^{2}.

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

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, -2 for b, and -3 for c in the quadratic formula.

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

Do the calculations.

$$t=\frac{2±4}{2}$$

Solve the equation t=\frac{2±4}{2} when ± is plus and when ± is minus.

$$t=3 t=-1$$

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

$$x=-\sqrt{3} x=\sqrt{3} x=-i x=i$$

Solve for x

$$x=-\sqrt{3}\approx -1.732050808$$

$$x=\sqrt{3}\approx 1.732050808$$

Steps Using the Quadratic Formula

Subtract 3 from both sides.

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

Substitute t for x^{2}.

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

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, -2 for b, and -3 for c in the quadratic formula.

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

Do the calculations.

$$t=\frac{2±4}{2}$$

Solve the equation t=\frac{2±4}{2} when ± is plus and when ± is minus.

$$t=3 t=-1$$

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

$$x=\sqrt{3} x=-\sqrt{3}$$