Solve 2sqrt{12}-6sqrt{1/3}+3sqrt{18}-sqrt{8}

$$2 \sqrt { 12 } - 6 \sqrt { \frac { 1 } { 3 } } + 3 \sqrt { 18 } - \sqrt { 8 }$$

Evaluate

$$\sqrt{2}\left(\sqrt{6}+7\right)\approx 13.363596552$$

Solution Steps

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}.

$$2\times 2\sqrt{3}-6\sqrt{\frac{1}{3}}+3\sqrt{18}-\sqrt{8}$$

Multiply 2 and 2 to get 4.

$$4\sqrt{3}-6\sqrt{\frac{1}{3}}+3\sqrt{18}-\sqrt{8}$$

Rewrite the square root of the division \sqrt{\frac{1}{3}} as the division of square roots \frac{\sqrt{1}}{\sqrt{3}}.

$$4\sqrt{3}-6\times \frac{\sqrt{1}}{\sqrt{3}}+3\sqrt{18}-\sqrt{8}$$

Calculate the square root of 1 and get 1.

$$4\sqrt{3}-6\times \frac{1}{\sqrt{3}}+3\sqrt{18}-\sqrt{8}$$

Rationalize the denominator of \frac{1}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$4\sqrt{3}-6\times \frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}+3\sqrt{18}-\sqrt{8}$$

The square of \sqrt{3} is 3.

$$4\sqrt{3}-6\times \frac{\sqrt{3}}{3}+3\sqrt{18}-\sqrt{8}$$

Cancel out 3, the greatest common factor in 6 and 3.

$$4\sqrt{3}-2\sqrt{3}+3\sqrt{18}-\sqrt{8}$$

Combine 4\sqrt{3} and -2\sqrt{3} to get 2\sqrt{3}.

$$2\sqrt{3}+3\sqrt{18}-\sqrt{8}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$2\sqrt{3}+3\times 3\sqrt{2}-\sqrt{8}$$

Multiply 3 and 3 to get 9.

$$2\sqrt{3}+9\sqrt{2}-\sqrt{8}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$2\sqrt{3}+9\sqrt{2}-2\sqrt{2}$$

Combine 9\sqrt{2} and -2\sqrt{2} to get 7\sqrt{2}.

$$2\sqrt{3}+7\sqrt{2}$$