Solve frac{2sqrt{3k^2left(k^2+1right)}}{4k^2+1}=1

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

Solve for k

$$k=-\frac{\sqrt{2}}{2}\approx -0.707106781$$

$$k=\frac{\sqrt{2}}{2}\approx 0.707106781$$

Solution Steps

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

$$2\sqrt{3k^{2}\left(k^{2}+1\right)}=4k^{2}+1$$

Use the distributive property to multiply 3k^{2} by k^{2}+1.

$$2\sqrt{3k^{4}+3k^{2}}=4k^{2}+1$$

Subtract 4k^{2} from both sides.

$$2\sqrt{3k^{4}+3k^{2}}-4k^{2}=1$$

Subtract -4k^{2} from both sides of the equation.

$$2\sqrt{3k^{4}+3k^{2}}=1+4k^{2}$$

Square both sides of the equation.

$$\left(2\sqrt{3k^{4}+3k^{2}}\right)^{2}=\left(4k^{2}+1\right)^{2}$$

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

$$2^{2}\left(\sqrt{3k^{4}+3k^{2}}\right)^{2}=\left(4k^{2}+1\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{3k^{4}+3k^{2}}\right)^{2}=\left(4k^{2}+1\right)^{2}$$

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

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

Use the distributive property to multiply 4 by 3k^{4}+3k^{2}.

$$12k^{4}+12k^{2}=\left(4k^{2}+1\right)^{2}$$

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

$$12k^{4}+12k^{2}=16\left(k^{2}\right)^{2}+8k^{2}+1$$

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

$$12k^{4}+12k^{2}=16k^{4}+8k^{2}+1$$

Subtract 16k^{4} from both sides.

$$12k^{4}+12k^{2}-16k^{4}=8k^{2}+1$$

Combine 12k^{4} and -16k^{4} to get -4k^{4}.

$$-4k^{4}+12k^{2}=8k^{2}+1$$

Subtract 8k^{2} from both sides.

$$-4k^{4}+12k^{2}-8k^{2}=1$$

Combine 12k^{2} and -8k^{2} to get 4k^{2}.

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

Subtract 1 from both sides.

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

Substitute t for k^{2}.

$$-4t^{2}+4t-1=0$$

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 -4 for a, 4 for b, and -1 for c in the quadratic formula.

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

Do the calculations.

$$t=\frac{-4±0}{-8}$$

Solutions are the same.

$$t=\frac{1}{2}$$

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

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

Substitute -\frac{\sqrt{2}}{2} for k in the equation \frac{2\sqrt{3k^{2}\left(k^{2}+1\right)}}{4k^{2}+1}=1.

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

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

$$1=1$$

Substitute \frac{\sqrt{2}}{2} for k in the equation \frac{2\sqrt{3k^{2}\left(k^{2}+1\right)}}{4k^{2}+1}=1.

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

Simplify. The value k=\frac{\sqrt{2}}{2} satisfies the equation.

$$1=1$$

List all solutions of 2\sqrt{3k^{4}+3k^{2}}=4k^{2}+1.

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