Solve 2sqrt{12}+6sqrt{1/3}-3sqrt{48}

$$2 \sqrt{ 12 } +6 \sqrt{ \frac{ 1 }{ 3 } } -3 \sqrt{ 48 }$$

Evaluate

$$-6\sqrt{3}\approx -10.392304845$$

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}.

$$2\times 2\sqrt{3}+6\sqrt{\frac{1}{3}}-3\sqrt{48}$$

Multiply 2 and 2 to get 4.

$$4\sqrt{3}+6\sqrt{\frac{1}{3}}-3\sqrt{48}$$

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

$$4\sqrt{3}+6\times \frac{\sqrt{1}}{\sqrt{3}}-3\sqrt{48}$$

Calculate the square root of 1 and get 1.

$$4\sqrt{3}+6\times \frac{1}{\sqrt{3}}-3\sqrt{48}$$

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

$$4\sqrt{3}+6\times \frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-3\sqrt{48}$$

The square of \sqrt{3} is 3.

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

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

$$4\sqrt{3}+2\sqrt{3}-3\sqrt{48}$$

Combine 4\sqrt{3} and 2\sqrt{3} to get 6\sqrt{3}.

$$6\sqrt{3}-3\sqrt{48}$$

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}.

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

Multiply -3 and 4 to get -12.

$$6\sqrt{3}-12\sqrt{3}$$

Combine 6\sqrt{3} and -12\sqrt{3} to get -6\sqrt{3}.

$$-6\sqrt{3}$$