Solve x^4=-x^2+12

$$x ^ { 4 } = - x ^ { 2 } + 12$$

Solve for x (complex solution)

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

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

$$x=-2i$$

$$x=2i$$

Steps Using the Quadratic Formula

Add x^{2} to both sides.

$$x^{4}+x^{2}=12$$

Subtract 12 from both sides.

$$x^{4}+x^{2}-12=0$$

Substitute t for x^{2}.

$$t^{2}+t-12=0$$

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 -12 for c in the quadratic formula.

$$t=\frac{-1±\sqrt{1^{2}-4\times 1\left(-12\right)}}{2}$$

Do the calculations.

$$t=\frac{-1±7}{2}$$

Solve the equation t=\frac{-1±7}{2} when ± is plus and when ± is minus.

$$t=3 t=-4$$

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

$$x=-\sqrt{3} x=\sqrt{3} x=-2i x=2i$$

Solve for x

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

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

Steps Using the Quadratic Formula

Add x^{2} to both sides.

$$x^{4}+x^{2}=12$$

Subtract 12 from both sides.

$$x^{4}+x^{2}-12=0$$

Substitute t for x^{2}.

$$t^{2}+t-12=0$$

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 -12 for c in the quadratic formula.

$$t=\frac{-1±\sqrt{1^{2}-4\times 1\left(-12\right)}}{2}$$

Do the calculations.

$$t=\frac{-1±7}{2}$$

Solve the equation t=\frac{-1±7}{2} when ± is plus and when ± is minus.

$$t=3 t=-4$$

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

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