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

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

Evaluate

$$\sqrt{3}\approx 1.732050808$$

Solution Steps

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

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

Consider \left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

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

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

The square of \sqrt{3} is 3.

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

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

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

Subtract 3 from 3 to get 0.

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

Cancel out 2 and 2.

$$\sqrt{3}$$