Solve 2/6+sqrt{3}

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

Evaluate

$$-\frac{2\sqrt{3}}{33}+\frac{4}{11}\approx 0.258663587$$

Solution Steps

Rationalize the denominator of \frac{2}{6+\sqrt{3}} by multiplying numerator and denominator by 6-\sqrt{3}.

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

Consider \left(6+\sqrt{3}\right)\left(6-\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}.

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

Square 6. Square \sqrt{3}.

$$\frac{2\left(6-\sqrt{3}\right)}{36-3}$$

Subtract 3 from 36 to get 33.

$$\frac{2\left(6-\sqrt{3}\right)}{33}$$

Use the distributive property to multiply 2 by 6-\sqrt{3}.

$$\frac{12-2\sqrt{3}}{33}$$