Solve 3/3-sqrt{3}

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

Evaluate

$$\frac{\sqrt{3}+3}{2}\approx 2.366025404$$

Solution Steps

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

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

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

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

Square 3. Square \sqrt{3}.

$$\frac{3\left(3+\sqrt{3}\right)}{9-3}$$

Subtract 3 from 9 to get 6.

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

Divide 3\left(3+\sqrt{3}\right) by 6 to get \frac{1}{2}\left(3+\sqrt{3}\right).

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

Use the distributive property to multiply \frac{1}{2} by 3+\sqrt{3}.

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

Multiply \frac{1}{2} and 3 to get \frac{3}{2}.

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