Solve sqrt{(6/sqrt{3})^2}+6

$$\sqrt { ( \frac { 6 } { \sqrt { 3 } } ) ^ { 2 } } + 6$$

Evaluate

$$2\sqrt{3}+6\approx 9.464101615$$

Solution Steps

Rationalize the denominator of \frac{6}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\sqrt{\left(\frac{6\sqrt{3}}{\left(\sqrt{3}\right)^{2}}\right)^{2}}+6$$

The square of \sqrt{3} is 3.

$$\sqrt{\left(\frac{6\sqrt{3}}{3}\right)^{2}}+6$$

Divide 6\sqrt{3} by 3 to get 2\sqrt{3}.

$$\sqrt{\left(2\sqrt{3}\right)^{2}}+6$$

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

$$\sqrt{2^{2}\left(\sqrt{3}\right)^{2}}+6$$

Calculate 2 to the power of 2 and get 4.

$$\sqrt{4\left(\sqrt{3}\right)^{2}}+6$$

The square of \sqrt{3} is 3.

$$\sqrt{4\times 3}+6$$

Multiply 4 and 3 to get 12.

$$\sqrt{12}+6$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$2\sqrt{3}+6$$