Solve 2sqrt{12}-sqrt{27}-6sqrt{4/3}+3sqrt{48}

$$2 \sqrt { 12 } - \sqrt { 27 } - 6 \sqrt { \frac { 4 } { 3 } } + 3 \sqrt { 48 }$$

Evaluate

$$9\sqrt{3}\approx 15.588457268$$

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}.

$$2\times 2\sqrt{3}-\sqrt{27}-6\sqrt{\frac{4}{3}}+3\sqrt{48}$$

Multiply 2 and 2 to get 4.

$$4\sqrt{3}-\sqrt{27}-6\sqrt{\frac{4}{3}}+3\sqrt{48}$$

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

$$4\sqrt{3}-3\sqrt{3}-6\sqrt{\frac{4}{3}}+3\sqrt{48}$$

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

$$\sqrt{3}-6\sqrt{\frac{4}{3}}+3\sqrt{48}$$

Rewrite the square root of the division \sqrt{\frac{4}{3}} as the division of square roots \frac{\sqrt{4}}{\sqrt{3}}.

$$\sqrt{3}-6\times \frac{\sqrt{4}}{\sqrt{3}}+3\sqrt{48}$$

Calculate the square root of 4 and get 2.

$$\sqrt{3}-6\times \frac{2}{\sqrt{3}}+3\sqrt{48}$$

Rationalize the denominator of \frac{2}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\sqrt{3}-6\times \frac{2\sqrt{3}}{\left(\sqrt{3}\right)^{2}}+3\sqrt{48}$$

The square of \sqrt{3} is 3.

$$\sqrt{3}-6\times \frac{2\sqrt{3}}{3}+3\sqrt{48}$$

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

$$\sqrt{3}-2\times 2\sqrt{3}+3\sqrt{48}$$

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

$$\sqrt{3}-2\times 2\sqrt{3}+3\times 4\sqrt{3}$$

Multiply 3 and 4 to get 12.

$$\sqrt{3}-2\times 2\sqrt{3}+12\sqrt{3}$$

Combine \sqrt{3} and 12\sqrt{3} to get 13\sqrt{3}.

$$13\sqrt{3}-2\times 2\sqrt{3}$$

Multiply -2 and 2 to get -4.

$$13\sqrt{3}-4\sqrt{3}$$

Combine 13\sqrt{3} and -4\sqrt{3} to get 9\sqrt{3}.

$$9\sqrt{3}$$