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

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

Evaluate

$$-\sqrt{3}-2\approx -3.732050808$$

Solution Steps

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

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

Consider \left(1-\sqrt{3}\right)\left(1+\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}.

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

Square 1. Square \sqrt{3}.

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

Subtract 3 from 1 to get -2.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Add 1 and 3 to get 4.

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

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

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