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

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

Evaluate

$$-5\sqrt{3}-5\approx -13.660254038$$

Solution Steps

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

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

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}.

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

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

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

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

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

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}.

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

Square 2. Square \sqrt{3}.

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

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Add 4 and 3 to get 7.

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

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

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

Subtract 7 from 2 to get -5.

$$-5-\sqrt{3}-4\sqrt{3}$$

Combine -\sqrt{3} and -4\sqrt{3} to get -5\sqrt{3}.

$$-5-5\sqrt{3}$$