Solve sqrt{2}*sqrt{6}+sqrt{1/18}-3sqrt{1/3}-sqrt{6}+sqrt{2}

$$\sqrt { 2 } \times \sqrt { 6 } + \sqrt { \frac { 1 } { 18 } } - 3 \sqrt { \frac { 1 } { 3 } } - \sqrt { 6 } + \sqrt { 2 }$$

Evaluate

$$\frac{7\sqrt{2}}{6}+\sqrt{3}-\sqrt{6}\approx 0.932476888$$

Solution Steps

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$\sqrt{2}\sqrt{2}\sqrt{3}+\sqrt{\frac{1}{18}}-3\sqrt{\frac{1}{3}}-\sqrt{6}+\sqrt{2}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$2\sqrt{3}+\sqrt{\frac{1}{18}}-3\sqrt{\frac{1}{3}}-\sqrt{6}+\sqrt{2}$$

Rewrite the square root of the division \sqrt{\frac{1}{18}} as the division of square roots \frac{\sqrt{1}}{\sqrt{18}}.

$$2\sqrt{3}+\frac{\sqrt{1}}{\sqrt{18}}-3\sqrt{\frac{1}{3}}-\sqrt{6}+\sqrt{2}$$

Calculate the square root of 1 and get 1.

$$2\sqrt{3}+\frac{1}{\sqrt{18}}-3\sqrt{\frac{1}{3}}-\sqrt{6}+\sqrt{2}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

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

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

$$2\sqrt{3}+\frac{\sqrt{2}}{3\left(\sqrt{2}\right)^{2}}-3\sqrt{\frac{1}{3}}-\sqrt{6}+\sqrt{2}$$

The square of \sqrt{2} is 2.

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

Multiply 3 and 2 to get 6.

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

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

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

Calculate the square root of 1 and get 1.

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

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

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

The square of \sqrt{3} is 3.

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

Cancel out 3 and 3.

$$2\sqrt{3}+\frac{\sqrt{2}}{6}-\sqrt{3}-\sqrt{6}+\sqrt{2}$$

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

$$\sqrt{3}+\frac{\sqrt{2}}{6}-\sqrt{6}+\sqrt{2}$$

Combine \frac{\sqrt{2}}{6} and \sqrt{2} to get \frac{7}{6}\sqrt{2}.

$$\sqrt{3}+\frac{7}{6}\sqrt{2}-\sqrt{6}$$