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

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

Evaluate

$$\frac{\sqrt{5}+3}{2}\approx 2.618033989$$

Solution Steps

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

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

Consider \left(\sqrt{5}-1\right)\left(\sqrt{5}+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}.

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

Square \sqrt{5}. Square 1.

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

Subtract 1 from 5 to get 4.

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

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

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

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

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

The square of \sqrt{5} is 5.

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

Add 5 and 1 to get 6.

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