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

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

Evaluate

$$\frac{10\sqrt{3}}{3}+8\approx 13.773502692$$

Short Solution Steps

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

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

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}.

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

Square 2. Square \sqrt{3}.

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

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

Multiply 2+\sqrt{3} and 2+\sqrt{3} to get \left(2+\sqrt{3}\right)^{2}.

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 2 and 3 to get 6.

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

Cancel out 6, the greatest common factor in 2 and 6.

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

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

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

The square of \sqrt{3} is 3.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2+\sqrt{3}\right)^{2}.

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

The square of \sqrt{3} is 3.

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

Add 4 and 3 to get 7.

$$7+4\sqrt{3}-\frac{2\sqrt{3}-3}{3}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 7+4\sqrt{3} times \frac{3}{3}.

$$\frac{3\left(7+4\sqrt{3}\right)}{3}-\frac{2\sqrt{3}-3}{3}$$

Since \frac{3\left(7+4\sqrt{3}\right)}{3} and \frac{2\sqrt{3}-3}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{3\left(7+4\sqrt{3}\right)-\left(2\sqrt{3}-3\right)}{3}$$

Do the multiplications in 3\left(7+4\sqrt{3}\right)-\left(2\sqrt{3}-3\right).

$$\frac{21+12\sqrt{3}-2\sqrt{3}+3}{3}$$

Do the calculations in 21+12\sqrt{3}-2\sqrt{3}+3.

$$\frac{24+10\sqrt{3}}{3}$$

Factor

$$\frac{2 {(5 \sqrt{3} + 12)}}{3} = 13.773502691896256$$