Solve (x)^4-15=(x)^2+13

$$( x ) ^ { 4 } - 15 = ( x ) ^ { 2 } + 13$$

Solve for x (complex solution)

$$x=\frac{i\sqrt{2\left(\sqrt{113}-1\right)}}{2}\approx 2.194327438i$$

$$x=-\frac{i\sqrt{2\left(\sqrt{113}-1\right)}}{2}\approx -0-2.194327438i$$

$$x = -\frac{\sqrt{2 {(\sqrt{113} + 1)}}}{2} \approx -2.411446227$$

$$x = \frac{\sqrt{2 {(\sqrt{113} + 1)}}}{2} \approx 2.411446227$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$x^{4}-15-x^{2}=13$$

Subtract 13 from both sides.

$$x^{4}-15-x^{2}-13=0$$

Subtract 13 from -15 to get -28.

$$x^{4}-28-x^{2}=0$$

Substitute t for x^{2}.

$$t^{2}-t-28=0$$

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 -28 for c in the quadratic formula.

$$t=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\left(-28\right)}}{2}$$

Do the calculations.

$$t=\frac{1±\sqrt{113}}{2}$$

Solve the equation t=\frac{1±\sqrt{113}}{2} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{113}+1}{2} t=\frac{1-\sqrt{113}}{2}$$

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

$$x=-\sqrt{\frac{\sqrt{113}+1}{2}} x=\sqrt{\frac{\sqrt{113}+1}{2}} x=-i\sqrt{-\frac{1-\sqrt{113}}{2}} x=i\sqrt{-\frac{1-\sqrt{113}}{2}}$$

Solve for x

$$x = -\frac{\sqrt{2 {(\sqrt{113} + 1)}}}{2} \approx -2.411446227$$

$$x = \frac{\sqrt{2 {(\sqrt{113} + 1)}}}{2} \approx 2.411446227$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$x^{4}-15-x^{2}=13$$

Subtract 13 from both sides.

$$x^{4}-15-x^{2}-13=0$$

Subtract 13 from -15 to get -28.

$$x^{4}-28-x^{2}=0$$

Substitute t for x^{2}.

$$t^{2}-t-28=0$$

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 -28 for c in the quadratic formula.

$$t=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\left(-28\right)}}{2}$$

Do the calculations.

$$t=\frac{1±\sqrt{113}}{2}$$

Solve the equation t=\frac{1±\sqrt{113}}{2} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{113}+1}{2} t=\frac{1-\sqrt{113}}{2}$$

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

$$x=\frac{\sqrt{2\sqrt{113}+2}}{2} x=-\frac{\sqrt{2\sqrt{113}+2}}{2}$$