Solve 400k^4=16k^4+32k^2+16

$$400 k ^ { 4 } = 16 k ^ { 4 } + 32 k ^ { 2 } + 16$$

Solve for k

$$k=-\frac{1}{2}=-0.5$$

$$k=\frac{1}{2}=0.5$$

Steps Using the Quadratic Formula

Subtract 16k^{4} from both sides.

$$400k^{4}-16k^{4}=32k^{2}+16$$

Combine 400k^{4} and -16k^{4} to get 384k^{4}.

$$384k^{4}=32k^{2}+16$$

Subtract 32k^{2} from both sides.

$$384k^{4}-32k^{2}=16$$

Subtract 16 from both sides.

$$384k^{4}-32k^{2}-16=0$$

Substitute t for k^{2}.

$$384t^{2}-32t-16=0$$

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 384 for a, -32 for b, and -16 for c in the quadratic formula.

$$t=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 384\left(-16\right)}}{2\times 384}$$

Do the calculations.

$$t=\frac{32±160}{768}$$

Solve the equation t=\frac{32±160}{768} when ± is plus and when ± is minus.

$$t=\frac{1}{4} t=-\frac{1}{6}$$

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

$$k=\frac{1}{2} k=-\frac{1}{2}$$