Solve 6x^4+4x^2=4

$$6 x ^ { 4 } + 4 x ^ { 2 } = 4$$

Solve for x (complex solution)

$$x=\frac{i\sqrt{3\left(\sqrt{7}+1\right)}}{3}\approx 1.102383979i$$

$$x=-\frac{i\sqrt{3\left(\sqrt{7}+1\right)}}{3}\approx -0-1.102383979i$$

$$x=-\frac{\sqrt{3\left(\sqrt{7}-1\right)}}{3}\approx -0.740664411$$

$$x=\frac{\sqrt{3\left(\sqrt{7}-1\right)}}{3}\approx 0.740664411$$

Steps Using the Quadratic Formula

Subtract 4 from both sides.

$$6x^{4}+4x^{2}-4=0$$

Substitute t for x^{2}.

$$6t^{2}+4t-4=0$$

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

$$t=\frac{-4±\sqrt{4^{2}-4\times 6\left(-4\right)}}{2\times 6}$$

Do the calculations.

$$t=\frac{-4±4\sqrt{7}}{12}$$

Solve the equation t=\frac{-4±4\sqrt{7}}{12} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{7}-1}{3} t=\frac{-\sqrt{7}-1}{3}$$

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

$$x=-\sqrt{\frac{\sqrt{7}-1}{3}} x=\sqrt{\frac{\sqrt{7}-1}{3}} x=-i\sqrt{\frac{\sqrt{7}+1}{3}} x=i\sqrt{\frac{\sqrt{7}+1}{3}}$$

Solve for x

$$x=-\frac{\sqrt{3\left(\sqrt{7}-1\right)}}{3}\approx -0.740664411$$

$$x=\frac{\sqrt{3\left(\sqrt{7}-1\right)}}{3}\approx 0.740664411$$

Steps Using the Quadratic Formula

Subtract 4 from both sides.

$$6x^{4}+4x^{2}-4=0$$

Substitute t for x^{2}.

$$6t^{2}+4t-4=0$$

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

$$t=\frac{-4±\sqrt{4^{2}-4\times 6\left(-4\right)}}{2\times 6}$$

Do the calculations.

$$t=\frac{-4±4\sqrt{7}}{12}$$

Solve the equation t=\frac{-4±4\sqrt{7}}{12} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{7}-1}{3} t=\frac{-\sqrt{7}-1}{3}$$

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

$$x=\sqrt{\frac{\sqrt{7}-1}{3}} x=-\sqrt{\frac{\sqrt{7}-1}{3}}$$