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

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

Evaluate

$$\frac{7\sqrt{3}}{6}-2\approx 0.020725942$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Multiply 2 and 3 to get 6.

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

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

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

Consider \left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square 2. Square \sqrt{3}.

$$\frac{\sqrt{3}}{6}-\frac{2-\sqrt{3}}{4-3}$$

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

To find the opposite of 2-\sqrt{3}, find the opposite of each term.

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

The opposite of -\sqrt{3} is \sqrt{3}.

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

Combine \frac{\sqrt{3}}{6} and \sqrt{3} to get \frac{7}{6}\sqrt{3}.

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

Factor

$$\frac{7 \sqrt{3} - 12}{6} = 0.02072594216369013$$