Solve 12K^2-1/12K^2+1=Ksqrt{3/3K^2+4}

$$\frac { 12 K ^ { 2 } - 1 } { 12 K ^ { 2 } + 1 } = K \sqrt { \frac { 3 } { 3 K ^ { 2 } + 4 } }$$

Solve for K

$$K=\frac{\sqrt{6}}{6}\approx 0.40824829$$

$$K=-\frac{\sqrt{2}}{6}\approx -0.23570226$$

Solution Steps

Square both sides of the equation.

$$\left(\frac{12K^{2}-1}{12K^{2}+1}\right)^{2}=\left(K\sqrt{\frac{3}{3K^{2}+4}}\right)^{2}$$

To raise \frac{12K^{2}-1}{12K^{2}+1} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(12K^{2}-1\right)^{2}}{\left(12K^{2}+1\right)^{2}}=\left(K\sqrt{\frac{3}{3K^{2}+4}}\right)^{2}$$

Expand \left(K\sqrt{\frac{3}{3K^{2}+4}}\right)^{2}.

$$\frac{\left(12K^{2}-1\right)^{2}}{\left(12K^{2}+1\right)^{2}}=K^{2}\left(\sqrt{\frac{3}{3K^{2}+4}}\right)^{2}$$

Calculate \sqrt{\frac{3}{3K^{2}+4}} to the power of 2 and get \frac{3}{3K^{2}+4}.

$$\frac{\left(12K^{2}-1\right)^{2}}{\left(12K^{2}+1\right)^{2}}=K^{2}\times \frac{3}{3K^{2}+4}$$

Express K^{2}\times \frac{3}{3K^{2}+4} as a single fraction.

$$\frac{\left(12K^{2}-1\right)^{2}}{\left(12K^{2}+1\right)^{2}}=\frac{K^{2}\times 3}{3K^{2}+4}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(12K^{2}-1\right)^{2}.

$$\frac{144\left(K^{2}\right)^{2}-24K^{2}+1}{\left(12K^{2}+1\right)^{2}}=\frac{K^{2}\times 3}{3K^{2}+4}$$

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

$$\frac{144K^{4}-24K^{2}+1}{\left(12K^{2}+1\right)^{2}}=\frac{K^{2}\times 3}{3K^{2}+4}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(12K^{2}+1\right)^{2}.

$$\frac{144K^{4}-24K^{2}+1}{144\left(K^{2}\right)^{2}+24K^{2}+1}=\frac{K^{2}\times 3}{3K^{2}+4}$$

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

$$\frac{144K^{4}-24K^{2}+1}{144K^{4}+24K^{2}+1}=\frac{K^{2}\times 3}{3K^{2}+4}$$

Multiply both sides of the equation by \left(3K^{2}+4\right)\left(12K^{2}+1\right)^{2}, the least common multiple of 144K^{4}+24K^{2}+1,3K^{2}+4.

$$\left(3K^{2}+4\right)\left(144K^{4}-24K^{2}+1\right)=\left(12K^{2}+1\right)^{2}K^{2}\times 3$$

Reorder the terms.

$$\left(3K^{2}+4\right)\left(144K^{4}-24K^{2}+1\right)=3K^{2}\left(12K^{2}+1\right)^{2}$$

Use the distributive property to multiply 3K^{2}+4 by 144K^{4}-24K^{2}+1 and combine like terms.

$$432K^{6}+504K^{4}-93K^{2}+4=3K^{2}\left(12K^{2}+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(12K^{2}+1\right)^{2}.

$$432K^{6}+504K^{4}-93K^{2}+4=3K^{2}\left(144\left(K^{2}\right)^{2}+24K^{2}+1\right)$$

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

$$432K^{6}+504K^{4}-93K^{2}+4=3K^{2}\left(144K^{4}+24K^{2}+1\right)$$

Use the distributive property to multiply 3K^{2} by 144K^{4}+24K^{2}+1.

$$432K^{6}+504K^{4}-93K^{2}+4=432K^{6}+72K^{4}+3K^{2}$$

Subtract 432K^{6} from both sides.

$$432K^{6}+504K^{4}-93K^{2}+4-432K^{6}=72K^{4}+3K^{2}$$

Combine 432K^{6} and -432K^{6} to get 0.

$$504K^{4}-93K^{2}+4=72K^{4}+3K^{2}$$

Subtract 72K^{4} from both sides.

$$504K^{4}-93K^{2}+4-72K^{4}=3K^{2}$$

Combine 504K^{4} and -72K^{4} to get 432K^{4}.

$$432K^{4}-93K^{2}+4=3K^{2}$$

Subtract 3K^{2} from both sides.

$$432K^{4}-93K^{2}+4-3K^{2}=0$$

Combine -93K^{2} and -3K^{2} to get -96K^{2}.

$$432K^{4}-96K^{2}+4=0$$

Substitute t for K^{2}.

$$432t^{2}-96t+4=0$$

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 432 for a, -96 for b, and 4 for c in the quadratic formula.

$$t=\frac{-\left(-96\right)±\sqrt{\left(-96\right)^{2}-4\times 432\times 4}}{2\times 432}$$

Do the calculations.

$$t=\frac{96±48}{864}$$

Solve the equation t=\frac{96±48}{864} when ± is plus and when ± is minus.

$$t=\frac{1}{6} t=\frac{1}{18}$$

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

$$K=\frac{\sqrt{6}}{6} K=-\frac{\sqrt{6}}{6} K=\frac{\sqrt{2}}{6} K=-\frac{\sqrt{2}}{6}$$

Substitute \frac{\sqrt{6}}{6} for K in the equation \frac{12K^{2}-1}{12K^{2}+1}=K\sqrt{\frac{3}{3K^{2}+4}}.

$$\frac{12\times \left(\frac{\sqrt{6}}{6}\right)^{2}-1}{12\times \left(\frac{\sqrt{6}}{6}\right)^{2}+1}=\frac{\sqrt{6}}{6}\sqrt{\frac{3}{3\times \left(\frac{\sqrt{6}}{6}\right)^{2}+4}}$$

Simplify. The value K=\frac{\sqrt{6}}{6} satisfies the equation.

$$\frac{1}{3}=\frac{1}{3}$$

Substitute -\frac{\sqrt{6}}{6} for K in the equation \frac{12K^{2}-1}{12K^{2}+1}=K\sqrt{\frac{3}{3K^{2}+4}}.

$$\frac{12\left(-\frac{\sqrt{6}}{6}\right)^{2}-1}{12\left(-\frac{\sqrt{6}}{6}\right)^{2}+1}=\left(-\frac{\sqrt{6}}{6}\right)\sqrt{\frac{3}{3\left(-\frac{\sqrt{6}}{6}\right)^{2}+4}}$$

Simplify. The value K=-\frac{\sqrt{6}}{6} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{1}{3}=-\frac{1}{3}$$

Substitute \frac{\sqrt{2}}{6} for K in the equation \frac{12K^{2}-1}{12K^{2}+1}=K\sqrt{\frac{3}{3K^{2}+4}}.

$$\frac{12\times \left(\frac{\sqrt{2}}{6}\right)^{2}-1}{12\times \left(\frac{\sqrt{2}}{6}\right)^{2}+1}=\frac{\sqrt{2}}{6}\sqrt{\frac{3}{3\times \left(\frac{\sqrt{2}}{6}\right)^{2}+4}}$$

Simplify. The value K=\frac{\sqrt{2}}{6} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-\frac{1}{5}=\frac{1}{5}$$

Substitute -\frac{\sqrt{2}}{6} for K in the equation \frac{12K^{2}-1}{12K^{2}+1}=K\sqrt{\frac{3}{3K^{2}+4}}.

$$\frac{12\left(-\frac{\sqrt{2}}{6}\right)^{2}-1}{12\left(-\frac{\sqrt{2}}{6}\right)^{2}+1}=\left(-\frac{\sqrt{2}}{6}\right)\sqrt{\frac{3}{3\left(-\frac{\sqrt{2}}{6}\right)^{2}+4}}$$

Simplify. The value K=-\frac{\sqrt{2}}{6} satisfies the equation.

$$-\frac{1}{5}=-\frac{1}{5}$$

List all solutions of \frac{12K^{2}-1}{12K^{2}+1}=\sqrt{\frac{3}{3K^{2}+4}}K.

$$K=\frac{\sqrt{6}}{6} K=-\frac{\sqrt{2}}{6}$$