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

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

Evaluate

$$14-8\sqrt{3}\approx 0.143593539$$

Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2}{2}.

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

Since \frac{2}{2} and \frac{\sqrt{3}}{2} have the same denominator, add them by adding their numerators.

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

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

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

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

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

Consider \left(2+\sqrt{3}\right)\left(2-\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}.

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

Square 2. Square \sqrt{3}.

$$\frac{\left(2-\sqrt{3}\right)\times 2\left(2-\sqrt{3}\right)}{4-3}$$

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

Use the distributive property to multiply 2-\sqrt{3} by 2.

$$\left(4-2\sqrt{3}\right)\left(2-\sqrt{3}\right)$$

Apply the distributive property by multiplying each term of 4-2\sqrt{3} by each term of 2-\sqrt{3}.

$$8-4\sqrt{3}-4\sqrt{3}+2\left(\sqrt{3}\right)^{2}$$

Combine -4\sqrt{3} and -4\sqrt{3} to get -8\sqrt{3}.

$$8-8\sqrt{3}+2\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$8-8\sqrt{3}+2\times 3$$

Multiply 2 and 3 to get 6.

$$8-8\sqrt{3}+6$$

Add 8 and 6 to get 14.

$$14-8\sqrt{3}$$