Solve sqrt{27}-6sqrt{1/3}+sqrt{48}

$$\sqrt { 27 } - 6 \sqrt { \frac { 1 } { 3 } } + \sqrt { 48 }$$

Evaluate

$$5\sqrt{3}\approx 8.660254038$$

Solution Steps

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}.

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

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

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

Calculate the square root of 1 and get 1.

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

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

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

The square of \sqrt{3} is 3.

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

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

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

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

$$\sqrt{3}+\sqrt{48}$$

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}.

$$\sqrt{3}+4\sqrt{3}$$

Combine \sqrt{3} and 4\sqrt{3} to get 5\sqrt{3}.

$$5\sqrt{3}$$