Solve frac{3-sqrt{7}}{3+sqrt{7}}+frac{3+sqrt{7}}{3-sqrt{7}}

$$\frac { 3 - \sqrt { 7 } } { 3 + \sqrt { 7 } } + \frac { 3 + \sqrt { 7 } } { 3 - \sqrt { 7 } }$$

Evaluate

$$16$$

Solution Steps

Rationalize the denominator of \frac{3-\sqrt{7}}{3+\sqrt{7}} by multiplying numerator and denominator by 3-\sqrt{7}.

$$\frac{\left(3-\sqrt{7}\right)\left(3-\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}+\frac{3+\sqrt{7}}{3-\sqrt{7}}$$

Consider \left(3+\sqrt{7}\right)\left(3-\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}.

$$\frac{\left(3-\sqrt{7}\right)\left(3-\sqrt{7}\right)}{3^{2}-\left(\sqrt{7}\right)^{2}}+\frac{3+\sqrt{7}}{3-\sqrt{7}}$$

Square 3. Square \sqrt{7}.

$$\frac{\left(3-\sqrt{7}\right)\left(3-\sqrt{7}\right)}{9-7}+\frac{3+\sqrt{7}}{3-\sqrt{7}}$$

Subtract 7 from 9 to get 2.

$$\frac{\left(3-\sqrt{7}\right)\left(3-\sqrt{7}\right)}{2}+\frac{3+\sqrt{7}}{3-\sqrt{7}}$$

Multiply 3-\sqrt{7} and 3-\sqrt{7} to get \left(3-\sqrt{7}\right)^{2}.

$$\frac{\left(3-\sqrt{7}\right)^{2}}{2}+\frac{3+\sqrt{7}}{3-\sqrt{7}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3-\sqrt{7}\right)^{2}.

$$\frac{9-6\sqrt{7}+\left(\sqrt{7}\right)^{2}}{2}+\frac{3+\sqrt{7}}{3-\sqrt{7}}$$

The square of \sqrt{7} is 7.

$$\frac{9-6\sqrt{7}+7}{2}+\frac{3+\sqrt{7}}{3-\sqrt{7}}$$

Add 9 and 7 to get 16.

$$\frac{16-6\sqrt{7}}{2}+\frac{3+\sqrt{7}}{3-\sqrt{7}}$$

Divide each term of 16-6\sqrt{7} by 2 to get 8-3\sqrt{7}.

$$8-3\sqrt{7}+\frac{3+\sqrt{7}}{3-\sqrt{7}}$$

Rationalize the denominator of \frac{3+\sqrt{7}}{3-\sqrt{7}} by multiplying numerator and denominator by 3+\sqrt{7}.

$$8-3\sqrt{7}+\frac{\left(3+\sqrt{7}\right)\left(3+\sqrt{7}\right)}{\left(3-\sqrt{7}\right)\left(3+\sqrt{7}\right)}$$

Consider \left(3-\sqrt{7}\right)\left(3+\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}.

$$8-3\sqrt{7}+\frac{\left(3+\sqrt{7}\right)\left(3+\sqrt{7}\right)}{3^{2}-\left(\sqrt{7}\right)^{2}}$$

Square 3. Square \sqrt{7}.

$$8-3\sqrt{7}+\frac{\left(3+\sqrt{7}\right)\left(3+\sqrt{7}\right)}{9-7}$$

Subtract 7 from 9 to get 2.

$$8-3\sqrt{7}+\frac{\left(3+\sqrt{7}\right)\left(3+\sqrt{7}\right)}{2}$$

Multiply 3+\sqrt{7} and 3+\sqrt{7} to get \left(3+\sqrt{7}\right)^{2}.

$$8-3\sqrt{7}+\frac{\left(3+\sqrt{7}\right)^{2}}{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+\sqrt{7}\right)^{2}.

$$8-3\sqrt{7}+\frac{9+6\sqrt{7}+\left(\sqrt{7}\right)^{2}}{2}$$

The square of \sqrt{7} is 7.

$$8-3\sqrt{7}+\frac{9+6\sqrt{7}+7}{2}$$

Add 9 and 7 to get 16.

$$8-3\sqrt{7}+\frac{16+6\sqrt{7}}{2}$$

Divide each term of 16+6\sqrt{7} by 2 to get 8+3\sqrt{7}.

$$8-3\sqrt{7}+8+3\sqrt{7}$$

Add 8 and 8 to get 16.

$$16-3\sqrt{7}+3\sqrt{7}$$

Combine -3\sqrt{7} and 3\sqrt{7} to get 0.

$$16$$

Factor

$$2^{4}$$