Solve frac{sqrt{4}}{sqrt{3}}-frac{sqrt{3}}{sqrt{4}}=

$$\frac{ \sqrt{ 4 } }{ \sqrt{ 3 } } - \frac{ \sqrt{ 3 } }{ \sqrt{ 4 } } =$$

Evaluate

$$\frac{\sqrt{3}}{6}\approx 0.288675135$$

Solution Steps

Calculate the square root of 4 and get 2.

$$\frac{2}{\sqrt{3}}-\frac{\sqrt{3}}{\sqrt{4}}$$

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

$$\frac{2\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-\frac{\sqrt{3}}{\sqrt{4}}$$

The square of \sqrt{3} is 3.

$$\frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{\sqrt{4}}$$

Calculate the square root of 4 and get 2.

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

Combine \frac{2\sqrt{3}}{3} and -\frac{\sqrt{3}}{2} to get \frac{1}{6}\sqrt{3}.

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