Solve 2/frac{1{sqrt{3}}+1}

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

Evaluate

$$3-\sqrt{3}\approx 1.267949192$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{3}{3}.

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

Since \frac{\sqrt{3}}{3} and \frac{3}{3} have the same denominator, add them by adding their numerators.

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

Divide 2 by \frac{\sqrt{3}+3}{3} by multiplying 2 by the reciprocal of \frac{\sqrt{3}+3}{3}.

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

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

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

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

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

Square \sqrt{3}. Square 3.

$$\frac{2\times 3\left(\sqrt{3}-3\right)}{3-9}$$

Subtract 9 from 3 to get -6.

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

Multiply 2 and 3 to get 6.

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

Cancel out -6 and -6.

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

To find the opposite of \sqrt{3}-3, find the opposite of each term.

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

The opposite of -3 is 3.

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