Solve 5sqrt{12}-9sqrt{1/3}+1/2sqrt{48}

$$5 \sqrt { 12 } - 9 \sqrt { \frac { 1 } { 3 } } + \frac { 1 } { 2 } \sqrt { 48 }$$

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}.

$$5\times 2\sqrt{3}-9\sqrt{\frac{1}{3}}+\frac{1}{2}\sqrt{48}$$

Multiply 5 and 2 to get 10.

$$10\sqrt{3}-9\sqrt{\frac{1}{3}}+\frac{1}{2}\sqrt{48}$$

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

$$10\sqrt{3}-9\times \frac{\sqrt{1}}{\sqrt{3}}+\frac{1}{2}\sqrt{48}$$

Calculate the square root of 1 and get 1.

$$10\sqrt{3}-9\times \frac{1}{\sqrt{3}}+\frac{1}{2}\sqrt{48}$$

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

$$10\sqrt{3}-9\times \frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}+\frac{1}{2}\sqrt{48}$$

The square of \sqrt{3} is 3.

$$10\sqrt{3}-9\times \frac{\sqrt{3}}{3}+\frac{1}{2}\sqrt{48}$$

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

$$10\sqrt{3}-3\sqrt{3}+\frac{1}{2}\sqrt{48}$$

Combine 10\sqrt{3} and -3\sqrt{3} to get 7\sqrt{3}.

$$7\sqrt{3}+\frac{1}{2}\sqrt{48}$$

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}.

$$7\sqrt{3}+\frac{1}{2}\times 4\sqrt{3}$$

Multiply \frac{1}{2} and 4 to get \frac{4}{2}.

$$7\sqrt{3}+\frac{4}{2}\sqrt{3}$$

Divide 4 by 2 to get 2.

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

Combine 7\sqrt{3} and 2\sqrt{3} to get 9\sqrt{3}.

$$9\sqrt{3}$$