Solve sqrt{1/3}*sqrt{6}-sqrt{8}

$$\sqrt { \frac { 1 } { 3 } } \times \sqrt { 6 } - \sqrt { 8 }$$

Evaluate

$$-\sqrt{2}\approx -1.414213562$$

Short Solution Steps

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

$$\frac{\sqrt{1}}{\sqrt{3}}\sqrt{6}-\sqrt{8}$$

Calculate the square root of 1 and get 1.

$$\frac{1}{\sqrt{3}}\sqrt{6}-\sqrt{8}$$

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

$$\frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}\sqrt{6}-\sqrt{8}$$

The square of \sqrt{3} is 3.

$$\frac{\sqrt{3}}{3}\sqrt{6}-\sqrt{8}$$

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

$$\frac{\sqrt{3}\sqrt{6}}{3}-\sqrt{8}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply -2\sqrt{2} times \frac{3}{3}.

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

Since \frac{\sqrt{3}\sqrt{6}}{3} and \frac{3\left(-2\right)\sqrt{2}}{3} have the same denominator, add them by adding their numerators.

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

Do the multiplications in \sqrt{3}\sqrt{6}+3\left(-2\right)\sqrt{2}.

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

Do the calculations in 3\sqrt{2}-6\sqrt{2}.

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

Cancel out 3 and 3.

$$-\sqrt{2}$$