Solve frac{6sqrt{k^2+k^4}}{3k^2+1}=frac{1/2sqrt{1+k^2}sqrt{48-288{k}^2}}{3{k}^2+1}

$$\frac{ 6 \sqrt{ { k }^{ 2 } + { k }^{ 4 } } }{ 3 { k }^{ 2 } +1 } = \frac{ \frac{ 1 }{ 2 } \sqrt{ 1+ { k }^{ 2 } } \sqrt{ 48-288 { k }^{ 2 } } }{ 3 { k }^{ 2 } +1 }$$

Solve for k

$$k=\frac{1}{3}\approx 0.333333333$$

$$k=-\frac{1}{3}\approx -0.333333333$$

Solution Steps

Multiply both sides of the equation by 3k^{2}+1.

$$6\sqrt{k^{2}+k^{4}}=\frac{1}{2}\sqrt{1+k^{2}}\sqrt{48-288k^{2}}$$

Subtract \frac{1}{2}\sqrt{1+k^{2}}\sqrt{48-288k^{2}} from both sides.

$$6\sqrt{k^{2}+k^{4}}-\frac{1}{2}\sqrt{1+k^{2}}\sqrt{48-288k^{2}}=0$$

Subtract -\frac{1}{2}\sqrt{1+k^{2}}\sqrt{48-288k^{2}} from both sides of the equation.

$$6\sqrt{k^{2}+k^{4}}=\frac{1}{2}\sqrt{1+k^{2}}\sqrt{48-288k^{2}}$$

Square both sides of the equation.

$$\left(6\sqrt{k^{2}+k^{4}}\right)^{2}=\left(\frac{1}{2}\sqrt{1+k^{2}}\sqrt{48-288k^{2}}\right)^{2}$$

Expand \left(6\sqrt{k^{2}+k^{4}}\right)^{2}.

$$6^{2}\left(\sqrt{k^{2}+k^{4}}\right)^{2}=\left(\frac{1}{2}\sqrt{1+k^{2}}\sqrt{48-288k^{2}}\right)^{2}$$

Calculate 6 to the power of 2 and get 36.

$$36\left(\sqrt{k^{2}+k^{4}}\right)^{2}=\left(\frac{1}{2}\sqrt{1+k^{2}}\sqrt{48-288k^{2}}\right)^{2}$$

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

$$36\left(k^{2}+k^{4}\right)=\left(\frac{1}{2}\sqrt{1+k^{2}}\sqrt{48-288k^{2}}\right)^{2}$$

Use the distributive property to multiply 36 by k^{2}+k^{4}.

$$36k^{2}+36k^{4}=\left(\frac{1}{2}\sqrt{1+k^{2}}\sqrt{48-288k^{2}}\right)^{2}$$

Expand \left(\frac{1}{2}\sqrt{1+k^{2}}\sqrt{48-288k^{2}}\right)^{2}.

$$36k^{2}+36k^{4}=\left(\frac{1}{2}\right)^{2}\left(\sqrt{1+k^{2}}\right)^{2}\left(\sqrt{48-288k^{2}}\right)^{2}$$

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$36k^{2}+36k^{4}=\frac{1}{4}\left(\sqrt{1+k^{2}}\right)^{2}\left(\sqrt{48-288k^{2}}\right)^{2}$$

Calculate \sqrt{1+k^{2}} to the power of 2 and get 1+k^{2}.

$$36k^{2}+36k^{4}=\frac{1}{4}\left(1+k^{2}\right)\left(\sqrt{48-288k^{2}}\right)^{2}$$

Calculate \sqrt{48-288k^{2}} to the power of 2 and get 48-288k^{2}.

$$36k^{2}+36k^{4}=\frac{1}{4}\left(1+k^{2}\right)\left(48-288k^{2}\right)$$

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

$$36k^{2}+36k^{4}=\left(\frac{1}{4}+\frac{1}{4}k^{2}\right)\left(48-288k^{2}\right)$$

Use the distributive property to multiply \frac{1}{4}+\frac{1}{4}k^{2} by 48-288k^{2} and combine like terms.

$$36k^{2}+36k^{4}=12-60k^{2}-72k^{4}$$

Subtract 12 from both sides.

$$36k^{2}+36k^{4}-12=-60k^{2}-72k^{4}$$

Add 60k^{2} to both sides.

$$36k^{2}+36k^{4}-12+60k^{2}=-72k^{4}$$

Combine 36k^{2} and 60k^{2} to get 96k^{2}.

$$96k^{2}+36k^{4}-12=-72k^{4}$$

Add 72k^{4} to both sides.

$$96k^{2}+36k^{4}-12+72k^{4}=0$$

Combine 36k^{4} and 72k^{4} to get 108k^{4}.

$$96k^{2}+108k^{4}-12=0$$

Substitute t for k^{2}.

$$108t^{2}+96t-12=0$$

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 108 for a, 96 for b, and -12 for c in the quadratic formula.

$$t=\frac{-96±\sqrt{96^{2}-4\times 108\left(-12\right)}}{2\times 108}$$

Do the calculations.

$$t=\frac{-96±120}{216}$$

Solve the equation t=\frac{-96±120}{216} when ± is plus and when ± is minus.

$$t=\frac{1}{9} t=-1$$

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

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

Substitute \frac{1}{3} for k in the equation \frac{6\sqrt{k^{2}+k^{4}}}{3k^{2}+1}=\frac{\frac{1}{2}\sqrt{1+k^{2}}\sqrt{48-288k^{2}}}{3k^{2}+1}.

$$\frac{6\sqrt{\left(\frac{1}{3}\right)^{2}+\left(\frac{1}{3}\right)^{4}}}{3\times \left(\frac{1}{3}\right)^{2}+1}=\frac{\frac{1}{2}\sqrt{1+\left(\frac{1}{3}\right)^{2}}\sqrt{48-288\times \left(\frac{1}{3}\right)^{2}}}{3\times \left(\frac{1}{3}\right)^{2}+1}$$

Simplify. The value k=\frac{1}{3} satisfies the equation.

$$\frac{1}{2}\times 10^{\frac{1}{2}}=\frac{1}{2}\times 10^{\frac{1}{2}}$$

Substitute -\frac{1}{3} for k in the equation \frac{6\sqrt{k^{2}+k^{4}}}{3k^{2}+1}=\frac{\frac{1}{2}\sqrt{1+k^{2}}\sqrt{48-288k^{2}}}{3k^{2}+1}.

$$\frac{6\sqrt{\left(-\frac{1}{3}\right)^{2}+\left(-\frac{1}{3}\right)^{4}}}{3\left(-\frac{1}{3}\right)^{2}+1}=\frac{\frac{1}{2}\sqrt{1+\left(-\frac{1}{3}\right)^{2}}\sqrt{48-288\left(-\frac{1}{3}\right)^{2}}}{3\left(-\frac{1}{3}\right)^{2}+1}$$

Simplify. The value k=-\frac{1}{3} satisfies the equation.

$$\frac{1}{2}\times 10^{\frac{1}{2}}=\frac{1}{2}\times 10^{\frac{1}{2}}$$

List all solutions of 6\sqrt{k^{4}+k^{2}}=\frac{\sqrt{48-288k^{2}}\sqrt{k^{2}+1}}{2}.

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

Solve for k (complex solution)

$$k=i$$

$$k=-i$$

$$k=-\frac{1}{3}\approx -0.333333333$$

$$k=\frac{1}{3}\approx 0.333333333$$