Solve x^4+x^2=4

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

Solve for x (complex solution)

$$x=\frac{i\sqrt{2\left(\sqrt{17}+1\right)}}{2}\approx 1.60048518i$$

$$x=-\frac{i\sqrt{2\left(\sqrt{17}+1\right)}}{2}\approx -0-1.60048518i$$

$$x = -\frac{\sqrt{2 {(\sqrt{17} - 1)}}}{2} \approx -1.249621068$$

$$x = \frac{\sqrt{2 {(\sqrt{17} - 1)}}}{2} \approx 1.249621068$$

Steps Using the Quadratic Formula

Subtract 4 from both sides.

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

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.

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

Do the calculations.

$$t=\frac{-1±\sqrt{17}}{2}$$

Solve the equation t=\frac{-1±\sqrt{17}}{2} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{17}-1}{2} t=\frac{-\sqrt{17}-1}{2}$$

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

$$x=-\sqrt{\frac{\sqrt{17}-1}{2}} x=\sqrt{\frac{\sqrt{17}-1}{2}} x=-i\sqrt{\frac{\sqrt{17}+1}{2}} x=i\sqrt{\frac{\sqrt{17}+1}{2}}$$

Solve for x

$$x = -\frac{\sqrt{2 {(\sqrt{17} - 1)}}}{2} \approx -1.249621068$$

$$x = \frac{\sqrt{2 {(\sqrt{17} - 1)}}}{2} \approx 1.249621068$$

Steps Using the Quadratic Formula

Subtract 4 from both sides.

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

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.

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

Do the calculations.

$$t=\frac{-1±\sqrt{17}}{2}$$

Solve the equation t=\frac{-1±\sqrt{17}}{2} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{17}-1}{2} t=\frac{-\sqrt{17}-1}{2}$$

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

$$x=\frac{\sqrt{2\sqrt{17}-2}}{2} x=-\frac{\sqrt{2\sqrt{17}-2}}{2}$$