Solve frac{5-sqrt{7}}{5+sqrt{7}}+frac{5+sqrt{7}}{5-sqrt{7}}

$$\frac { 5 - \sqrt { 7 } } { 5 + \sqrt { 7 } } + \frac { 5 + \sqrt { 7 } } { 5 - \sqrt { 7 } }$$

Evaluate

$$\frac{32}{9}\approx 3.555555556$$

Short Solution Steps

Rationalize the denominator of \frac{5-\sqrt{7}}{5+\sqrt{7}} by multiplying numerator and denominator by 5-\sqrt{7}.

$$\frac{\left(5-\sqrt{7}\right)\left(5-\sqrt{7}\right)}{\left(5+\sqrt{7}\right)\left(5-\sqrt{7}\right)}+\frac{5+\sqrt{7}}{5-\sqrt{7}}$$

Consider \left(5+\sqrt{7}\right)\left(5-\sqrt{7}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(5-\sqrt{7}\right)\left(5-\sqrt{7}\right)}{5^{2}-\left(\sqrt{7}\right)^{2}}+\frac{5+\sqrt{7}}{5-\sqrt{7}}$$

Square 5. Square \sqrt{7}.

$$\frac{\left(5-\sqrt{7}\right)\left(5-\sqrt{7}\right)}{25-7}+\frac{5+\sqrt{7}}{5-\sqrt{7}}$$

Subtract 7 from 25 to get 18.

$$\frac{\left(5-\sqrt{7}\right)\left(5-\sqrt{7}\right)}{18}+\frac{5+\sqrt{7}}{5-\sqrt{7}}$$

Multiply 5-\sqrt{7} and 5-\sqrt{7} to get \left(5-\sqrt{7}\right)^{2}.

$$\frac{\left(5-\sqrt{7}\right)^{2}}{18}+\frac{5+\sqrt{7}}{5-\sqrt{7}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-\sqrt{7}\right)^{2}.

$$\frac{25-10\sqrt{7}+\left(\sqrt{7}\right)^{2}}{18}+\frac{5+\sqrt{7}}{5-\sqrt{7}}$$

The square of \sqrt{7} is 7.

$$\frac{25-10\sqrt{7}+7}{18}+\frac{5+\sqrt{7}}{5-\sqrt{7}}$$

Add 25 and 7 to get 32.

$$\frac{32-10\sqrt{7}}{18}+\frac{5+\sqrt{7}}{5-\sqrt{7}}$$

Rationalize the denominator of \frac{5+\sqrt{7}}{5-\sqrt{7}} by multiplying numerator and denominator by 5+\sqrt{7}.

$$\frac{32-10\sqrt{7}}{18}+\frac{\left(5+\sqrt{7}\right)\left(5+\sqrt{7}\right)}{\left(5-\sqrt{7}\right)\left(5+\sqrt{7}\right)}$$

Consider \left(5-\sqrt{7}\right)\left(5+\sqrt{7}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{32-10\sqrt{7}}{18}+\frac{\left(5+\sqrt{7}\right)\left(5+\sqrt{7}\right)}{5^{2}-\left(\sqrt{7}\right)^{2}}$$

Square 5. Square \sqrt{7}.

$$\frac{32-10\sqrt{7}}{18}+\frac{\left(5+\sqrt{7}\right)\left(5+\sqrt{7}\right)}{25-7}$$

Subtract 7 from 25 to get 18.

$$\frac{32-10\sqrt{7}}{18}+\frac{\left(5+\sqrt{7}\right)\left(5+\sqrt{7}\right)}{18}$$

Multiply 5+\sqrt{7} and 5+\sqrt{7} to get \left(5+\sqrt{7}\right)^{2}.

$$\frac{32-10\sqrt{7}}{18}+\frac{\left(5+\sqrt{7}\right)^{2}}{18}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(5+\sqrt{7}\right)^{2}.

$$\frac{32-10\sqrt{7}}{18}+\frac{25+10\sqrt{7}+\left(\sqrt{7}\right)^{2}}{18}$$

The square of \sqrt{7} is 7.

$$\frac{32-10\sqrt{7}}{18}+\frac{25+10\sqrt{7}+7}{18}$$

Add 25 and 7 to get 32.

$$\frac{32-10\sqrt{7}}{18}+\frac{32+10\sqrt{7}}{18}$$

Since \frac{32-10\sqrt{7}}{18} and \frac{32+10\sqrt{7}}{18} have the same denominator, add them by adding their numerators.

$$\frac{32-10\sqrt{7}+32+10\sqrt{7}}{18}$$

Do the calculations in 32-10\sqrt{7}+32+10\sqrt{7}.

$$\frac{64}{18}$$

Reduce the fraction \frac{64}{18} to lowest terms by extracting and canceling out 2.

$$\frac{32}{9}$$

Factor

$$\frac{2 ^ {5}}{3 ^ {2}} = 3\frac{5}{9} = 3.5555555555555554$$