Solve frac{4+sqrt{5}}{4-sqrt{5}}+frac{4-sqrt{5}}{4+sqrt{5}}

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

Evaluate

$$\frac{42}{11}\approx 3.818181818$$

Short Solution Steps

Rationalize the denominator of \frac{4+\sqrt{5}}{4-\sqrt{5}} by multiplying numerator and denominator by 4+\sqrt{5}.

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

Consider \left(4-\sqrt{5}\right)\left(4+\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}.

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

Square 4. Square \sqrt{5}.

$$\frac{\left(4+\sqrt{5}\right)\left(4+\sqrt{5}\right)}{16-5}+\frac{4-\sqrt{5}}{4+\sqrt{5}}$$

Subtract 5 from 16 to get 11.

$$\frac{\left(4+\sqrt{5}\right)\left(4+\sqrt{5}\right)}{11}+\frac{4-\sqrt{5}}{4+\sqrt{5}}$$

Multiply 4+\sqrt{5} and 4+\sqrt{5} to get \left(4+\sqrt{5}\right)^{2}.

$$\frac{\left(4+\sqrt{5}\right)^{2}}{11}+\frac{4-\sqrt{5}}{4+\sqrt{5}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4+\sqrt{5}\right)^{2}.

$$\frac{16+8\sqrt{5}+\left(\sqrt{5}\right)^{2}}{11}+\frac{4-\sqrt{5}}{4+\sqrt{5}}$$

The square of \sqrt{5} is 5.

$$\frac{16+8\sqrt{5}+5}{11}+\frac{4-\sqrt{5}}{4+\sqrt{5}}$$

Add 16 and 5 to get 21.

$$\frac{21+8\sqrt{5}}{11}+\frac{4-\sqrt{5}}{4+\sqrt{5}}$$

Rationalize the denominator of \frac{4-\sqrt{5}}{4+\sqrt{5}} by multiplying numerator and denominator by 4-\sqrt{5}.

$$\frac{21+8\sqrt{5}}{11}+\frac{\left(4-\sqrt{5}\right)\left(4-\sqrt{5}\right)}{\left(4+\sqrt{5}\right)\left(4-\sqrt{5}\right)}$$

Consider \left(4+\sqrt{5}\right)\left(4-\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}.

$$\frac{21+8\sqrt{5}}{11}+\frac{\left(4-\sqrt{5}\right)\left(4-\sqrt{5}\right)}{4^{2}-\left(\sqrt{5}\right)^{2}}$$

Square 4. Square \sqrt{5}.

$$\frac{21+8\sqrt{5}}{11}+\frac{\left(4-\sqrt{5}\right)\left(4-\sqrt{5}\right)}{16-5}$$

Subtract 5 from 16 to get 11.

$$\frac{21+8\sqrt{5}}{11}+\frac{\left(4-\sqrt{5}\right)\left(4-\sqrt{5}\right)}{11}$$

Multiply 4-\sqrt{5} and 4-\sqrt{5} to get \left(4-\sqrt{5}\right)^{2}.

$$\frac{21+8\sqrt{5}}{11}+\frac{\left(4-\sqrt{5}\right)^{2}}{11}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-\sqrt{5}\right)^{2}.

$$\frac{21+8\sqrt{5}}{11}+\frac{16-8\sqrt{5}+\left(\sqrt{5}\right)^{2}}{11}$$

The square of \sqrt{5} is 5.

$$\frac{21+8\sqrt{5}}{11}+\frac{16-8\sqrt{5}+5}{11}$$

Add 16 and 5 to get 21.

$$\frac{21+8\sqrt{5}}{11}+\frac{21-8\sqrt{5}}{11}$$

Since \frac{21+8\sqrt{5}}{11} and \frac{21-8\sqrt{5}}{11} have the same denominator, add them by adding their numerators.

$$\frac{21+8\sqrt{5}+21-8\sqrt{5}}{11}$$

Do the calculations in 21+8\sqrt{5}+21-8\sqrt{5}.

$$\frac{42}{11}$$

Factor

$$\frac{2 \cdot 3 \cdot 7}{11} = 3\frac{9}{11} = 3.8181818181818183$$