Solve frac{1-sqrt{5}}{1+sqrt{5}}

$$\frac { 1 - \sqrt { 5 } } { 1 + \sqrt { 5 } }$$

Evaluate

$$\frac{\sqrt{5}-3}{2}\approx -0.381966011$$

Solution Steps

Rationalize the denominator of \frac{1-\sqrt{5}}{1+\sqrt{5}} by multiplying numerator and denominator by 1-\sqrt{5}.

$$\frac{\left(1-\sqrt{5}\right)\left(1-\sqrt{5}\right)}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}$$

Consider \left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(1-\sqrt{5}\right)\left(1-\sqrt{5}\right)}{1^{2}-\left(\sqrt{5}\right)^{2}}$$

Square 1. Square \sqrt{5}.

$$\frac{\left(1-\sqrt{5}\right)\left(1-\sqrt{5}\right)}{1-5}$$

Subtract 5 from 1 to get -4.

$$\frac{\left(1-\sqrt{5}\right)\left(1-\sqrt{5}\right)}{-4}$$

Multiply 1-\sqrt{5} and 1-\sqrt{5} to get \left(1-\sqrt{5}\right)^{2}.

$$\frac{\left(1-\sqrt{5}\right)^{2}}{-4}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-\sqrt{5}\right)^{2}.

$$\frac{1-2\sqrt{5}+\left(\sqrt{5}\right)^{2}}{-4}$$

The square of \sqrt{5} is 5.

$$\frac{1-2\sqrt{5}+5}{-4}$$

Add 1 and 5 to get 6.

$$\frac{6-2\sqrt{5}}{-4}$$