Solve frac{2sqrt{5}+1}{2sqrt{5}-1}

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

Evaluate

$$\frac{4\sqrt{5}+21}{19}\approx 1.576014311$$

Solution Steps

Rationalize the denominator of \frac{2\sqrt{5}+1}{2\sqrt{5}-1} by multiplying numerator and denominator by 2\sqrt{5}+1.

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

Consider \left(2\sqrt{5}-1\right)\left(2\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}.

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

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

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

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

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

The square of \sqrt{5} is 5.

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

Multiply 4 and 5 to get 20.

$$\frac{20+4\sqrt{5}+1}{\left(2\sqrt{5}\right)^{2}-1^{2}}$$

Add 20 and 1 to get 21.

$$\frac{21+4\sqrt{5}}{\left(2\sqrt{5}\right)^{2}-1^{2}}$$

Expand \left(2\sqrt{5}\right)^{2}.

$$\frac{21+4\sqrt{5}}{2^{2}\left(\sqrt{5}\right)^{2}-1^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{21+4\sqrt{5}}{4\left(\sqrt{5}\right)^{2}-1^{2}}$$

The square of \sqrt{5} is 5.

$$\frac{21+4\sqrt{5}}{4\times 5-1^{2}}$$

Multiply 4 and 5 to get 20.

$$\frac{21+4\sqrt{5}}{20-1^{2}}$$

Calculate 1 to the power of 2 and get 1.

$$\frac{21+4\sqrt{5}}{20-1}$$

Subtract 1 from 20 to get 19.

$$\frac{21+4\sqrt{5}}{19}$$