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

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

Evaluate

$$\frac{1}{2}=0.5$$

Short Solution Steps

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

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

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}.

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

Square 5. Square \sqrt{5}.

$$\frac{5-\sqrt{5}}{25-5}+\frac{1}{5-\sqrt{5}}$$

Subtract 5 from 25 to get 20.

$$\frac{5-\sqrt{5}}{20}+\frac{1}{5-\sqrt{5}}$$

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

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

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}.

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

Square 5. Square \sqrt{5}.

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

Subtract 5 from 25 to get 20.

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

Since \frac{5-\sqrt{5}}{20} and \frac{5+\sqrt{5}}{20} have the same denominator, add them by adding their numerators.

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

Do the calculations in 5-\sqrt{5}+5+\sqrt{5}.

$$\frac{10}{20}$$

Reduce the fraction \frac{10}{20} to lowest terms by extracting and canceling out 10.

$$\frac{1}{2}$$

Factor

$$\frac{1}{2} = 0.5$$