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

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

Evaluate

$$2-\sqrt{3}\approx 0.267949192$$

Solution Steps

Divide \frac{1}{2} by \frac{2+\sqrt{3}}{2} by multiplying \frac{1}{2} by the reciprocal of \frac{2+\sqrt{3}}{2}.

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

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

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

Rationalize the denominator of \frac{2}{4+2\sqrt{3}} by multiplying numerator and denominator by 4-2\sqrt{3}.

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

Consider \left(4+2\sqrt{3}\right)\left(4-2\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}.

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

Calculate 4 to the power of 2 and get 16.

$$\frac{2\left(4-2\sqrt{3}\right)}{16-\left(2\sqrt{3}\right)^{2}}$$

Expand \left(2\sqrt{3}\right)^{2}.

$$\frac{2\left(4-2\sqrt{3}\right)}{16-2^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{2\left(4-2\sqrt{3}\right)}{16-4\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{2\left(4-2\sqrt{3}\right)}{16-4\times 3}$$

Multiply 4 and 3 to get 12.

$$\frac{2\left(4-2\sqrt{3}\right)}{16-12}$$

Subtract 12 from 16 to get 4.

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

Divide 2\left(4-2\sqrt{3}\right) by 4 to get \frac{1}{2}\left(4-2\sqrt{3}\right).

$$\frac{1}{2}\left(4-2\sqrt{3}\right)$$

Use the distributive property to multiply \frac{1}{2} by 4-2\sqrt{3}.

$$\frac{1}{2}\times 4+\frac{1}{2}\left(-2\right)\sqrt{3}$$

Multiply \frac{1}{2} and 4 to get \frac{4}{2}.

$$\frac{4}{2}+\frac{1}{2}\left(-2\right)\sqrt{3}$$

Divide 4 by 2 to get 2.

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

Multiply \frac{1}{2} and -2 to get \frac{-2}{2}.

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

Divide -2 by 2 to get -1.

$$2-\sqrt{3}$$