Solve sqrt{3/2}-sqrt{1/18}

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

Evaluate

$$\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{6}\approx 0.989042611$$

Short Solution Steps

Rewrite the square root of the division \sqrt{\frac{3}{2}} as the division of square roots \frac{\sqrt{3}}{\sqrt{2}}.

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

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

$$\frac{\sqrt{3}\sqrt{2}}{\left(\sqrt{2}\right)^{2}}-\sqrt{\frac{1}{18}}$$

The square of \sqrt{2} is 2.

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

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

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

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

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

Calculate the square root of 1 and get 1.

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

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}.

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

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

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

The square of \sqrt{2} is 2.

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

Multiply 3 and 2 to get 6.

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

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and 6 is 6. Multiply \frac{\sqrt{6}}{2} times \frac{3}{3}.

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

Since \frac{3\sqrt{6}}{6} and \frac{\sqrt{2}}{6} have the same denominator, subtract them by subtracting their numerators.

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