Solve frac{3sqrt{3}}{2-sqrt{3}}

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

Evaluate

$$6\sqrt{3}+9\approx 19.392304845$$

Solution Steps

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

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

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}.

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

Square 2. Square \sqrt{3}.

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

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

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

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

The square of \sqrt{3} is 3.

$$6\sqrt{3}+3\times 3$$

Multiply 3 and 3 to get 9.

$$6\sqrt{3}+9$$