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

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

Evaluate

$$8\sqrt{3}\approx 13.856406461$$

Solution Steps

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

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

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}.

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

Square 2. Square \sqrt{3}.

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

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

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

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

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

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

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}.

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

Square 2. Square \sqrt{3}.

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

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Add 4 and 3 to get 7.

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

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

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

The square of \sqrt{3} is 3.

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

Add 4 and 3 to get 7.

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

To find the opposite of 7-4\sqrt{3}, find the opposite of each term.

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

The opposite of -4\sqrt{3} is 4\sqrt{3}.

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

Subtract 7 from 7 to get 0.

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

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

$$8\sqrt{3}$$