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

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

Evaluate

$$12$$

Solution Steps

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

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

Consider \left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}+\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}.

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

Square \sqrt{7}. Square \sqrt{5}.

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

Subtract 5 from 7 to get 2.

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

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

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

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

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

The square of \sqrt{7} is 7.

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

To multiply \sqrt{7} and \sqrt{5}, multiply the numbers under the square root.

$$\frac{7+2\sqrt{35}+\left(\sqrt{5}\right)^{2}}{2}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$$

The square of \sqrt{5} is 5.

$$\frac{7+2\sqrt{35}+5}{2}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$$

Add 7 and 5 to get 12.

$$\frac{12+2\sqrt{35}}{2}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$$

Divide each term of 12+2\sqrt{35} by 2 to get 6+\sqrt{35}.

$$6+\sqrt{35}+\frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}+\sqrt{5}}$$

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

$$6+\sqrt{35}+\frac{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}{\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}$$

Consider \left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\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}.

$$6+\sqrt{35}+\frac{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}{\left(\sqrt{7}\right)^{2}-\left(\sqrt{5}\right)^{2}}$$

Square \sqrt{7}. Square \sqrt{5}.

$$6+\sqrt{35}+\frac{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}{7-5}$$

Subtract 5 from 7 to get 2.

$$6+\sqrt{35}+\frac{\left(\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)}{2}$$

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

$$6+\sqrt{35}+\frac{\left(\sqrt{7}-\sqrt{5}\right)^{2}}{2}$$

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

$$6+\sqrt{35}+\frac{\left(\sqrt{7}\right)^{2}-2\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^{2}}{2}$$

The square of \sqrt{7} is 7.

$$6+\sqrt{35}+\frac{7-2\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^{2}}{2}$$

To multiply \sqrt{7} and \sqrt{5}, multiply the numbers under the square root.

$$6+\sqrt{35}+\frac{7-2\sqrt{35}+\left(\sqrt{5}\right)^{2}}{2}$$

The square of \sqrt{5} is 5.

$$6+\sqrt{35}+\frac{7-2\sqrt{35}+5}{2}$$

Add 7 and 5 to get 12.

$$6+\sqrt{35}+\frac{12-2\sqrt{35}}{2}$$

Divide each term of 12-2\sqrt{35} by 2 to get 6-\sqrt{35}.

$$6+\sqrt{35}+6-\sqrt{35}$$

Add 6 and 6 to get 12.

$$12+\sqrt{35}-\sqrt{35}$$

Combine \sqrt{35} and -\sqrt{35} to get 0.

$$12$$

Factor

$$2^{2}\times 3$$