Solve 3x^4+x^2=4

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

Solve for x (complex solution)

$$x=1$$

$$x=-1$$

$$x=-\frac{2\sqrt{3}i}{3}\approx -0-1.154700538i$$

$$x=\frac{2\sqrt{3}i}{3}\approx 1.154700538i$$

Steps Using the Quadratic Formula

Subtract 4 from both sides.

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

Substitute t for x^{2}.

$$3t^{2}+t-4=0$$

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

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

Do the calculations.

$$t=\frac{-1±7}{6}$$

Solve the equation t=\frac{-1±7}{6} when ± is plus and when ± is minus.

$$t=1 t=-\frac{4}{3}$$

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

$$x=-1 x=1 x=-\frac{2\sqrt{3}i}{3} x=\frac{2\sqrt{3}i}{3}$$

Solve for x

$$x=-1$$

$$x=1$$

Steps Using the Quadratic Formula

Subtract 4 from both sides.

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

Substitute t for x^{2}.

$$3t^{2}+t-4=0$$

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

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

Do the calculations.

$$t=\frac{-1±7}{6}$$

Solve the equation t=\frac{-1±7}{6} when ± is plus and when ± is minus.

$$t=1 t=-\frac{4}{3}$$

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

$$x=1 x=-1$$