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

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

Evaluate

$$\sqrt{3}+16\approx 17.732050808$$

Solution Steps

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

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

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}.

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

Square 2. Square \sqrt{3}.

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

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

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

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

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

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

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}.

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

Square 2. Square \sqrt{3}.

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

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

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

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

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

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

Consider \left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

Multiply \sqrt{3}+1 and \sqrt{3}+1 to get \left(\sqrt{3}+1\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3}+1\right)^{2}.

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

Divide each term of 4+2\sqrt{3} by 2 to get 2+\sqrt{3}.

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

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

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

The square of \sqrt{3} is 3.

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

Add 4 and 3 to get 7.

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

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

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

The square of \sqrt{3} is 3.

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

Add 4 and 3 to get 7.

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

Add 7 and 7 to get 14.

$$14+4\sqrt{3}-4\sqrt{3}+2+\sqrt{3}$$

Combine 4\sqrt{3} and -4\sqrt{3} to get 0.

$$14+2+\sqrt{3}$$

Add 14 and 2 to get 16.

$$16+\sqrt{3}$$