Solve frac{6+sqrt{3}}{6-sqrt{3}}

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

Evaluate

$$\frac{4\sqrt{3}+13}{11}\approx 1.811654839$$

Solution Steps

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

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

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}.

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

Square 6. Square \sqrt{3}.

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

Subtract 3 from 36 to get 33.

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

Multiply 6+\sqrt{3} and 6+\sqrt{3} to get \left(6+\sqrt{3}\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(6+\sqrt{3}\right)^{2}.

$$\frac{36+12\sqrt{3}+\left(\sqrt{3}\right)^{2}}{33}$$

The square of \sqrt{3} is 3.

$$\frac{36+12\sqrt{3}+3}{33}$$

Add 36 and 3 to get 39.

$$\frac{39+12\sqrt{3}}{33}$$