Solve (1+k^2)^2=16k^2

$$( 1 + k ^ { 2 } ) ^ { 2 } = 16 k ^ { 2 }$$

Solve for k (complex solution)

$$k=2-\sqrt{3}\approx 0.267949192$$

$$k=\sqrt{3}-2\approx -0.267949192$$

$$k=-\left(\sqrt{3}+2\right)\approx -3.732050808$$

$$k=\sqrt{3}+2\approx 3.732050808$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+k^{2}\right)^{2}.

$$1+2k^{2}+\left(k^{2}\right)^{2}=16k^{2}$$

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

$$1+2k^{2}+k^{4}=16k^{2}$$

Subtract 16k^{2} from both sides.

$$1+2k^{2}+k^{4}-16k^{2}=0$$

Combine 2k^{2} and -16k^{2} to get -14k^{2}.

$$1-14k^{2}+k^{4}=0$$

Substitute t for k^{2}.

$$t^{2}-14t+1=0$$

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, -14 for b, and 1 for c in the quadratic formula.

$$t=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$t=\frac{14±8\sqrt{3}}{2}$$

Solve the equation t=\frac{14±8\sqrt{3}}{2} when ± is plus and when ± is minus.

$$t=4\sqrt{3}+7 t=7-4\sqrt{3}$$

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

$$k=-\left(\sqrt{3}+2\right) k=\sqrt{3}+2 k=-\left(2-\sqrt{3}\right) k=2-\sqrt{3}$$

Solve for k

$$k=\sqrt{3}-2\approx -0.267949192$$

$$k=2-\sqrt{3}\approx 0.267949192$$

$$k=\sqrt{3}+2\approx 3.732050808$$

$$k=-\sqrt{3}-2\approx -3.732050808$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(1+k^{2}\right)^{2}.

$$1+2k^{2}+\left(k^{2}\right)^{2}=16k^{2}$$

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

$$1+2k^{2}+k^{4}=16k^{2}$$

Subtract 16k^{2} from both sides.

$$1+2k^{2}+k^{4}-16k^{2}=0$$

Combine 2k^{2} and -16k^{2} to get -14k^{2}.

$$1-14k^{2}+k^{4}=0$$

Substitute t for k^{2}.

$$t^{2}-14t+1=0$$

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, -14 for b, and 1 for c in the quadratic formula.

$$t=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$t=\frac{14±8\sqrt{3}}{2}$$

Solve the equation t=\frac{14±8\sqrt{3}}{2} when ± is plus and when ± is minus.

$$t=4\sqrt{3}+7 t=7-4\sqrt{3}$$

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

$$k=\sqrt{3}+2 k=-\left(\sqrt{3}+2\right) k=2-\sqrt{3} k=-\left(2-\sqrt{3}\right)$$