Solve 3(-16k/4k^2+1)^2=32/4k^2+1

$$3 ( \frac { - 16 k } { 4 k ^ { 2 } + 1 } ) ^ { 2 } = \frac { 32 } { 4 k ^ { 2 } + 1 }$$

Solve for k

$$k=\frac{\sqrt{5}}{10}\approx 0.223606798$$

$$k=-\frac{\sqrt{5}}{10}\approx -0.223606798$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4k^{2}+1.

$$3\times \left(\frac{-16k}{4k^{2}+1}\right)^{2}\left(4k^{2}+1\right)=32$$

To raise \frac{-16k}{4k^{2}+1} to a power, raise both numerator and denominator to the power and then divide.

$$3\times \frac{\left(-16k\right)^{2}}{\left(4k^{2}+1\right)^{2}}\left(4k^{2}+1\right)=32$$

Express 3\times \frac{\left(-16k\right)^{2}}{\left(4k^{2}+1\right)^{2}} as a single fraction.

$$\frac{3\left(-16k\right)^{2}}{\left(4k^{2}+1\right)^{2}}\left(4k^{2}+1\right)=32$$

Express \frac{3\left(-16k\right)^{2}}{\left(4k^{2}+1\right)^{2}}\left(4k^{2}+1\right) as a single fraction.

$$\frac{3\left(-16k\right)^{2}\left(4k^{2}+1\right)}{\left(4k^{2}+1\right)^{2}}=32$$

Expand \left(-16k\right)^{2}.

$$\frac{3\left(-16\right)^{2}k^{2}\left(4k^{2}+1\right)}{\left(4k^{2}+1\right)^{2}}=32$$

Calculate -16 to the power of 2 and get 256.

$$\frac{3\times 256k^{2}\left(4k^{2}+1\right)}{\left(4k^{2}+1\right)^{2}}=32$$

Multiply 3 and 256 to get 768.

$$\frac{768k^{2}\left(4k^{2}+1\right)}{\left(4k^{2}+1\right)^{2}}=32$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4k^{2}+1\right)^{2}.

$$\frac{768k^{2}\left(4k^{2}+1\right)}{16\left(k^{2}\right)^{2}+8k^{2}+1}=32$$

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

$$\frac{768k^{2}\left(4k^{2}+1\right)}{16k^{4}+8k^{2}+1}=32$$

Subtract 32 from both sides.

$$\frac{768k^{2}\left(4k^{2}+1\right)}{16k^{4}+8k^{2}+1}-32=0$$

Use the distributive property to multiply 768k^{2} by 4k^{2}+1.

$$\frac{3072k^{4}+768k^{2}}{16k^{4}+8k^{2}+1}-32=0$$

Factor 16k^{4}+8k^{2}+1.

$$\frac{3072k^{4}+768k^{2}}{\left(4k^{2}+1\right)^{2}}-32=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 32 times \frac{\left(4k^{2}+1\right)^{2}}{\left(4k^{2}+1\right)^{2}}.

$$\frac{3072k^{4}+768k^{2}}{\left(4k^{2}+1\right)^{2}}-\frac{32\left(4k^{2}+1\right)^{2}}{\left(4k^{2}+1\right)^{2}}=0$$

Since \frac{3072k^{4}+768k^{2}}{\left(4k^{2}+1\right)^{2}} and \frac{32\left(4k^{2}+1\right)^{2}}{\left(4k^{2}+1\right)^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3072k^{4}+768k^{2}-32\left(4k^{2}+1\right)^{2}}{\left(4k^{2}+1\right)^{2}}=0$$

Do the multiplications in 3072k^{4}+768k^{2}-32\left(4k^{2}+1\right)^{2}.

$$\frac{3072k^{4}+768k^{2}-512k^{4}-256k^{2}-32}{\left(4k^{2}+1\right)^{2}}=0$$

Combine like terms in 3072k^{4}+768k^{2}-512k^{4}-256k^{2}-32.

$$\frac{2560k^{4}+512k^{2}-32}{\left(4k^{2}+1\right)^{2}}=0$$

Multiply both sides of the equation by \left(4k^{2}+1\right)^{2}.

$$2560k^{4}+512k^{2}-32=0$$

Substitute t for k^{2}.

$$2560t^{2}+512t-32=0$$

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 2560 for a, 512 for b, and -32 for c in the quadratic formula.

$$t=\frac{-512±\sqrt{512^{2}-4\times 2560\left(-32\right)}}{2\times 2560}$$

Do the calculations.

$$t=\frac{-512±768}{5120}$$

Solve the equation t=\frac{-512±768}{5120} when ± is plus and when ± is minus.

$$t=\frac{1}{20} t=-\frac{1}{4}$$

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

$$k=\frac{\sqrt{5}}{10} k=-\frac{\sqrt{5}}{10}$$