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

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

Evaluate

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

Solution Steps

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

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

Consider \left(5+\sqrt{5}\right)\left(5-\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}.

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

Square 5. Square \sqrt{5}.

$$\frac{\left(5-\sqrt{5}\right)\left(5-\sqrt{5}\right)}{25-5}$$

Subtract 5 from 25 to get 20.

$$\frac{\left(5-\sqrt{5}\right)\left(5-\sqrt{5}\right)}{20}$$

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

$$\frac{\left(5-\sqrt{5}\right)^{2}}{20}$$

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

$$\frac{25-10\sqrt{5}+\left(\sqrt{5}\right)^{2}}{20}$$

The square of \sqrt{5} is 5.

$$\frac{25-10\sqrt{5}+5}{20}$$

Add 25 and 5 to get 30.

$$\frac{30-10\sqrt{5}}{20}$$

Factor

$$\frac{3 - \sqrt{5}}{2} = 0.3819660112501051$$