Solve (x^4)^2=16-15(x^2)^2

$$( x ^ { 4 } ) ^ { 2 } = 16 - 15 ( x ^ { 2 } ) ^ { 2 }$$

Solve for x (complex solution)

$$x\in 1,-1,i,-i,\sqrt{2}\left(-1-i\right),\sqrt{2}\left(1-i\right),\sqrt{2}\left(-1+i\right),\sqrt{2}\left(1+i\right)$$

Steps Using the Quadratic Formula

To raise a power to another power, multiply the exponents. Multiply 4 and 2 to get 8.

$$x^{8}=16-15\left(x^{2}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$x^{8}=16-15x^{4}$$

Subtract 16 from both sides.

$$x^{8}-16=-15x^{4}$$

Add 15x^{4} to both sides.

$$x^{8}-16+15x^{4}=0$$

Substitute t for x^{4}.

$$t^{2}+15t-16=0$$

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, 15 for b, and -16 for c in the quadratic formula.

$$t=\frac{-15±\sqrt{15^{2}-4\times 1\left(-16\right)}}{2}$$

Do the calculations.

$$t=\frac{-15±17}{2}$$

Solve the equation t=\frac{-15±17}{2} when ± is plus and when ± is minus.

$$t=1 t=-16$$

Since x=t^{4}, the solutions are obtained by solving the equation for each t.

$$x=-1 x=-i x=i x=1 x=\sqrt{2}\left(-1-i\right) x=\sqrt{2}\left(1+i\right) x=\sqrt{2}\left(-1+i\right) x=\sqrt{2}\left(1-i\right)$$

Solve for x

$$x=-1$$

$$x=1$$

Steps Using the Quadratic Formula

To raise a power to another power, multiply the exponents. Multiply 4 and 2 to get 8.

$$x^{8}=16-15\left(x^{2}\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$x^{8}=16-15x^{4}$$

Subtract 16 from both sides.

$$x^{8}-16=-15x^{4}$$

Add 15x^{4} to both sides.

$$x^{8}-16+15x^{4}=0$$

Substitute t for x^{4}.

$$t^{2}+15t-16=0$$

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, 15 for b, and -16 for c in the quadratic formula.

$$t=\frac{-15±\sqrt{15^{2}-4\times 1\left(-16\right)}}{2}$$

Do the calculations.

$$t=\frac{-15±17}{2}$$

Solve the equation t=\frac{-15±17}{2} when ± is plus and when ± is minus.

$$t=1 t=-16$$

Since x=t^{4}, the solutions are obtained by evaluating x=±\sqrt[4]{t} for positive t.

$$x=1 x=-1$$