Solve (3)3sqrt{48}-9sqrt{1/3}+3sqrt{12}

$$( 3 ) 3 \sqrt { 48 } - 9 \sqrt { \frac { 1 } { 3 } } + 3 \sqrt { 12 }$$

Evaluate

$$39\sqrt{3}\approx 67.549981495$$

Solution Steps

Multiply 3 and 3 to get 9.

$$9\sqrt{48}-9\sqrt{\frac{1}{3}}+3\sqrt{12}$$

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}.

$$9\times 4\sqrt{3}-9\sqrt{\frac{1}{3}}+3\sqrt{12}$$

Multiply 9 and 4 to get 36.

$$36\sqrt{3}-9\sqrt{\frac{1}{3}}+3\sqrt{12}$$

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

$$36\sqrt{3}-9\times \frac{\sqrt{1}}{\sqrt{3}}+3\sqrt{12}$$

Calculate the square root of 1 and get 1.

$$36\sqrt{3}-9\times \frac{1}{\sqrt{3}}+3\sqrt{12}$$

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

$$36\sqrt{3}-9\times \frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}+3\sqrt{12}$$

The square of \sqrt{3} is 3.

$$36\sqrt{3}-9\times \frac{\sqrt{3}}{3}+3\sqrt{12}$$

Cancel out 3, the greatest common factor in 9 and 3.

$$36\sqrt{3}-3\sqrt{3}+3\sqrt{12}$$

Combine 36\sqrt{3} and -3\sqrt{3} to get 33\sqrt{3}.

$$33\sqrt{3}+3\sqrt{12}$$

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}.

$$33\sqrt{3}+3\times 2\sqrt{3}$$

Multiply 3 and 2 to get 6.

$$33\sqrt{3}+6\sqrt{3}$$

Combine 33\sqrt{3} and 6\sqrt{3} to get 39\sqrt{3}.

$$39\sqrt{3}$$