Solve x^2-x^3x=4

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

Solve for x (complex solution)

$$x=\sqrt{2}e^{\frac{\arctan(\sqrt{15})i+2\pi i}{2}}\approx -1.118033989-0.866025404i$$

$$x=\sqrt{2}e^{\frac{\arctan(\sqrt{15})i}{2}}\approx 1.118033989+0.866025404i$$

$$x=\sqrt{2}e^{\frac{-\arctan(\sqrt{15})i+2\pi i}{2}}\approx -1.118033989+0.866025404i$$

$$x=\sqrt{2}e^{-\frac{\arctan(\sqrt{15})i}{2}}\approx 1.118033989-0.866025404i$$

Steps Using the Quadratic Formula

To multiply powers of the same base, add their exponents. Add 3 and 1 to get 4.

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

Subtract 4 from both sides.

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

Substitute t for x^{2}.

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

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

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

Do the calculations.

$$t=\frac{-1±\sqrt{-15}}{-2}$$

Solve the equation t=\frac{-1±\sqrt{-15}}{-2} when ± is plus and when ± is minus.

$$t=\frac{-\sqrt{15}i+1}{2} t=\frac{1+\sqrt{15}i}{2}$$

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

$$x=\sqrt{2}e^{-\frac{\arctan(\sqrt{15})i}{2}} x=\sqrt{2}e^{\frac{-\arctan(\sqrt{15})i+2\pi i}{2}} x=\sqrt{2}e^{\frac{\arctan(\sqrt{15})i+2\pi i}{2}} x=\sqrt{2}e^{\frac{\arctan(\sqrt{15})i}{2}}$$