Solve sqrt{48}-frac{sqrt{27}}{2}+1/sqrt{12}4

$$\sqrt { 48 } - \frac { \sqrt { 27 } } { 2 } + \frac { 1 } { \sqrt { 12 } } 4$$

Evaluate

$$\frac{19\sqrt{3}}{6}\approx 5.484827557$$

Solution Steps

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}.

$$4\sqrt{3}-\frac{\sqrt{27}}{2}+4\times \frac{1}{\sqrt{12}}$$

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}.

$$4\sqrt{3}-\frac{3\sqrt{3}}{2}+4\times \frac{1}{\sqrt{12}}$$

Combine 4\sqrt{3} and -\frac{3\sqrt{3}}{2} to get \frac{5}{2}\sqrt{3}.

$$\frac{5}{2}\sqrt{3}+4\times \frac{1}{\sqrt{12}}$$

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}.

$$\frac{5}{2}\sqrt{3}+4\times \frac{1}{2\sqrt{3}}$$

Rationalize the denominator of \frac{1}{2\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{5}{2}\sqrt{3}+4\times \frac{\sqrt{3}}{2\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{5}{2}\sqrt{3}+4\times \frac{\sqrt{3}}{2\times 3}$$

Multiply 2 and 3 to get 6.

$$\frac{5}{2}\sqrt{3}+4\times \frac{\sqrt{3}}{6}$$

Express 4\times \frac{\sqrt{3}}{6} as a single fraction.

$$\frac{5}{2}\sqrt{3}+\frac{4\sqrt{3}}{6}$$

Combine \frac{5}{2}\sqrt{3} and \frac{4\sqrt{3}}{6} to get \frac{19}{6}\sqrt{3}.

$$\frac{19}{6}\sqrt{3}$$