Solve c^2=4sqrt{c^2+1}

$$c ^ { 2 } = 4 \sqrt { c ^ { 2 } + 1 }$$

Solve for c

$$c=2\sqrt{\sqrt{5}+2}\approx 4.116342055$$

$$c=-2\sqrt{\sqrt{5}+2}\approx -4.116342055$$

Solution Steps

Square both sides of the equation.

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

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

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

Expand \left(4\sqrt{c^{2}+1}\right)^{2}.

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

Calculate 4 to the power of 2 and get 16.

$$c^{4}=16\left(\sqrt{c^{2}+1}\right)^{2}$$

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

$$c^{4}=16\left(c^{2}+1\right)$$

Use the distributive property to multiply 16 by c^{2}+1.

$$c^{4}=16c^{2}+16$$

Subtract 16c^{2} from both sides.

$$c^{4}-16c^{2}=16$$

Subtract 16 from both sides.

$$c^{4}-16c^{2}-16=0$$

Substitute t for c^{2}.

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

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

$$t=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 1\left(-16\right)}}{2}$$

Do the calculations.

$$t=\frac{16±8\sqrt{5}}{2}$$

Solve the equation t=\frac{16±8\sqrt{5}}{2} when ± is plus and when ± is minus.

$$t=4\sqrt{5}+8 t=8-4\sqrt{5}$$

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

$$c=2\sqrt{\sqrt{5}+2} c=-2\sqrt{\sqrt{5}+2}$$

Substitute 2\sqrt{\sqrt{5}+2} for c in the equation c^{2}=4\sqrt{c^{2}+1}.

$$\left(2\sqrt{\sqrt{5}+2}\right)^{2}=4\sqrt{\left(2\sqrt{\sqrt{5}+2}\right)^{2}+1}$$

Simplify. The value c=2\sqrt{\sqrt{5}+2} satisfies the equation.

$$4\times 5^{\frac{1}{2}}+8=4\times 5^{\frac{1}{2}}+8$$

Substitute -2\sqrt{\sqrt{5}+2} for c in the equation c^{2}=4\sqrt{c^{2}+1}.

$$\left(-2\sqrt{\sqrt{5}+2}\right)^{2}=4\sqrt{\left(-2\sqrt{\sqrt{5}+2}\right)^{2}+1}$$

Simplify. The value c=-2\sqrt{\sqrt{5}+2} satisfies the equation.

$$4\times 5^{\frac{1}{2}}+8=4\times 5^{\frac{1}{2}}+8$$

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

$$c=2\sqrt{\sqrt{5}+2} c=-2\sqrt{\sqrt{5}+2}$$