Solve frac{1+sqrt{3}}{3+sqrt{3}}

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

Evaluate

$$\frac{\sqrt{3}}{3}\approx 0.577350269$$

Solution Steps

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

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

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}.

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

Square 3. Square \sqrt{3}.

$$\frac{\left(1+\sqrt{3}\right)\left(3-\sqrt{3}\right)}{9-3}$$

Subtract 3 from 9 to get 6.

$$\frac{\left(1+\sqrt{3}\right)\left(3-\sqrt{3}\right)}{6}$$

Apply the distributive property by multiplying each term of 1+\sqrt{3} by each term of 3-\sqrt{3}.

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

Combine -\sqrt{3} and 3\sqrt{3} to get 2\sqrt{3}.

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

The square of \sqrt{3} is 3.

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

Subtract 3 from 3 to get 0.

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

Divide 2\sqrt{3} by 6 to get \frac{1}{3}\sqrt{3}.

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