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

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

Evaluate

$$2\sqrt{3}+3\approx 6.464101615$$

Solution Steps

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

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

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}.

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 3 to get 6.

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

Combine 3\sqrt{3} and \sqrt{3} to get 4\sqrt{3}.

$$\frac{4\sqrt{3}+6}{2}$$

Divide each term of 4\sqrt{3}+6 by 2 to get 2\sqrt{3}+3.

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

Factor

$$\sqrt{3} {(\sqrt{3} + 2)} = 6.464101615$$