Solve c=2sqrt{16-{left(sqrt{3}-frac{sqrt{3}}{2}cright)}^2}

$$c = 2 \sqrt{ 16- { \left( \sqrt{ 3 } - \frac{ \sqrt{ 3 } }{ 2 } c \right) }^{ 2 } }$$

Solve for c

$$c = \frac{\sqrt{61} + 3}{2} \approx 5.405124838$$

Solution Steps

Square both sides of the equation.

$$c^{2}=\left(2\sqrt{16-\left(\sqrt{3}-\frac{\sqrt{3}}{2}c\right)^{2}}\right)^{2}$$

Express \frac{\sqrt{3}}{2}c as a single fraction.

$$c^{2}=\left(2\sqrt{16-\left(\sqrt{3}-\frac{\sqrt{3}c}{2}\right)^{2}}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3}-\frac{\sqrt{3}c}{2}\right)^{2}.

$$c^{2}=\left(2\sqrt{16-\left(\left(\sqrt{3}\right)^{2}+2\sqrt{3}\left(-\frac{\sqrt{3}c}{2}\right)+\left(-\frac{\sqrt{3}c}{2}\right)^{2}\right)}\right)^{2}$$

The square of \sqrt{3} is 3.

$$c^{2}=\left(2\sqrt{16-\left(3+2\sqrt{3}\left(-\frac{\sqrt{3}c}{2}\right)+\left(-\frac{\sqrt{3}c}{2}\right)^{2}\right)}\right)^{2}$$

Express 2\left(-\frac{\sqrt{3}c}{2}\right) as a single fraction.

$$c^{2}=\left(2\sqrt{16-\left(3+\frac{-2\sqrt{3}c}{2}\sqrt{3}+\left(-\frac{\sqrt{3}c}{2}\right)^{2}\right)}\right)^{2}$$

Cancel out 2 and 2.

$$c^{2}=\left(2\sqrt{16-\left(3-\sqrt{3}c\sqrt{3}+\left(-\frac{\sqrt{3}c}{2}\right)^{2}\right)}\right)^{2}$$

Calculate -\frac{\sqrt{3}c}{2} to the power of 2 and get \left(\frac{\sqrt{3}c}{2}\right)^{2}.

$$c^{2}=\left(2\sqrt{16-\left(3-\sqrt{3}c\sqrt{3}+\left(\frac{\sqrt{3}c}{2}\right)^{2}\right)}\right)^{2}$$

To find the opposite of 3-\sqrt{3}c\sqrt{3}+\left(\frac{\sqrt{3}c}{2}\right)^{2}, find the opposite of each term.

$$c^{2}=\left(2\sqrt{16-3+\sqrt{3}c\sqrt{3}-\left(\frac{\sqrt{3}c}{2}\right)^{2}}\right)^{2}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$c^{2}=\left(2\sqrt{16-3+3c-\left(\frac{\sqrt{3}c}{2}\right)^{2}}\right)^{2}$$

Subtract 3 from 16 to get 13.

$$c^{2}=\left(2\sqrt{13+3c-\left(\frac{\sqrt{3}c}{2}\right)^{2}}\right)^{2}$$

To raise \frac{\sqrt{3}c}{2} to a power, raise both numerator and denominator to the power and then divide.

$$c^{2}=\left(2\sqrt{13+3c-\frac{\left(\sqrt{3}c\right)^{2}}{2^{2}}}\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 13+3c times \frac{2^{2}}{2^{2}}.

$$c^{2}=\left(2\sqrt{\frac{\left(13+3c\right)\times 2^{2}}{2^{2}}-\frac{\left(\sqrt{3}c\right)^{2}}{2^{2}}}\right)^{2}$$

Since \frac{\left(13+3c\right)\times 2^{2}}{2^{2}} and \frac{\left(\sqrt{3}c\right)^{2}}{2^{2}} have the same denominator, subtract them by subtracting their numerators.

$$c^{2}=\left(2\sqrt{\frac{\left(13+3c\right)\times 2^{2}-\left(\sqrt{3}c\right)^{2}}{2^{2}}}\right)^{2}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3c times \frac{2^{2}}{2^{2}}.

$$c^{2}=\left(2\sqrt{13+\frac{3c\times 2^{2}}{2^{2}}-\frac{\left(\sqrt{3}c\right)^{2}}{2^{2}}}\right)^{2}$$

Since \frac{3c\times 2^{2}}{2^{2}} and \frac{\left(\sqrt{3}c\right)^{2}}{2^{2}} have the same denominator, subtract them by subtracting their numerators.

$$c^{2}=\left(2\sqrt{13+\frac{3c\times 2^{2}-\left(\sqrt{3}c\right)^{2}}{2^{2}}}\right)^{2}$$

Expand \left(2\sqrt{13+\frac{3c\times 2^{2}-\left(\sqrt{3}c\right)^{2}}{2^{2}}}\right)^{2}.

$$c^{2}=2^{2}\left(\sqrt{13+\frac{3c\times 2^{2}-\left(\sqrt{3}c\right)^{2}}{2^{2}}}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$c^{2}=4\left(\sqrt{13+\frac{3c\times 2^{2}-\left(\sqrt{3}c\right)^{2}}{2^{2}}}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$c^{2}=4\left(\sqrt{13+\frac{3c\times 4-\left(\sqrt{3}c\right)^{2}}{2^{2}}}\right)^{2}$$

Multiply 3 and 4 to get 12.

$$c^{2}=4\left(\sqrt{13+\frac{12c-\left(\sqrt{3}c\right)^{2}}{2^{2}}}\right)^{2}$$

Expand \left(\sqrt{3}c\right)^{2}.

$$c^{2}=4\left(\sqrt{13+\frac{12c-\left(\sqrt{3}\right)^{2}c^{2}}{2^{2}}}\right)^{2}$$

The square of \sqrt{3} is 3.

$$c^{2}=4\left(\sqrt{13+\frac{12c-3c^{2}}{2^{2}}}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$c^{2}=4\left(\sqrt{13+\frac{12c-3c^{2}}{4}}\right)^{2}$$

Calculate \sqrt{13+\frac{12c-3c^{2}}{4}} to the power of 2 and get 13+\frac{12c-3c^{2}}{4}.

$$c^{2}=4\left(13+\frac{12c-3c^{2}}{4}\right)$$

Use the distributive property to multiply 4 by 13+\frac{12c-3c^{2}}{4}.

$$c^{2}=52+4\times \frac{12c-3c^{2}}{4}$$

Express 4\times \frac{12c-3c^{2}}{4} as a single fraction.

$$c^{2}=52+\frac{4\left(12c-3c^{2}\right)}{4}$$

Cancel out 4 and 4.

$$c^{2}=52+12c-3c^{2}$$

Subtract 52 from both sides.

$$c^{2}-52=12c-3c^{2}$$

Subtract 12c from both sides.

$$c^{2}-52-12c=-3c^{2}$$

Add 3c^{2} to both sides.

$$c^{2}-52-12c+3c^{2}=0$$

Combine c^{2} and 3c^{2} to get 4c^{2}.

$$4c^{2}-52-12c=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$4c^{2}-12c-52=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -12 for b, and -52 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 4\left(-52\right)}}{2\times 4}$$

Square -12.

$$c=\frac{-\left(-12\right)±\sqrt{144-4\times 4\left(-52\right)}}{2\times 4}$$

Multiply -4 times 4.

$$c=\frac{-\left(-12\right)±\sqrt{144-16\left(-52\right)}}{2\times 4}$$

Multiply -16 times -52.

$$c=\frac{-\left(-12\right)±\sqrt{144+832}}{2\times 4}$$

Add 144 to 832.

$$c=\frac{-\left(-12\right)±\sqrt{976}}{2\times 4}$$

Take the square root of 976.

$$c=\frac{-\left(-12\right)±4\sqrt{61}}{2\times 4}$$

The opposite of -12 is 12.

$$c=\frac{12±4\sqrt{61}}{2\times 4}$$

Multiply 2 times 4.

$$c=\frac{12±4\sqrt{61}}{8}$$

Now solve the equation c=\frac{12±4\sqrt{61}}{8} when ± is plus. Add 12 to 4\sqrt{61}.

$$c=\frac{4\sqrt{61}+12}{8}$$

Divide 12+4\sqrt{61} by 8.

$$c=\frac{\sqrt{61}+3}{2}$$

Now solve the equation c=\frac{12±4\sqrt{61}}{8} when ± is minus. Subtract 4\sqrt{61} from 12.

$$c=\frac{12-4\sqrt{61}}{8}$$

Divide 12-4\sqrt{61} by 8.

$$c=\frac{3-\sqrt{61}}{2}$$

The equation is now solved.

$$c=\frac{\sqrt{61}+3}{2} c=\frac{3-\sqrt{61}}{2}$$

Substitute \frac{\sqrt{61}+3}{2} for c in the equation c=2\sqrt{16-\left(\sqrt{3}-\frac{\sqrt{3}}{2}c\right)^{2}}.

$$\frac{\sqrt{61}+3}{2}=2\sqrt{16-\left(\sqrt{3}-\frac{\sqrt{3}}{2}\times \frac{\sqrt{61}+3}{2}\right)^{2}}$$

Simplify. The value c=\frac{\sqrt{61}+3}{2} satisfies the equation.

$$\frac{1}{2}\times 61^{\frac{1}{2}}+\frac{3}{2}=\frac{1}{2}\times 61^{\frac{1}{2}}+\frac{3}{2}$$

Substitute \frac{3-\sqrt{61}}{2} for c in the equation c=2\sqrt{16-\left(\sqrt{3}-\frac{\sqrt{3}}{2}c\right)^{2}}.

$$\frac{3-\sqrt{61}}{2}=2\sqrt{16-\left(\sqrt{3}-\frac{\sqrt{3}}{2}\times \frac{3-\sqrt{61}}{2}\right)^{2}}$$

Simplify. The value c=\frac{3-\sqrt{61}}{2} does not satisfy the equation because the left and the right hand side have opposite signs.

$$\frac{3}{2}-\frac{1}{2}\times 61^{\frac{1}{2}}=\frac{1}{2}\times 61^{\frac{1}{2}}-\frac{3}{2}$$

Equation c=2\sqrt{16-\left(\sqrt{3}-\frac{\sqrt{3}}{2}c\right)^{2}} has a unique solution.

$$c=\frac{\sqrt{61}+3}{2}$$