Solve frac{sqrt{5}-sqrt{3}}{sqrt{5}+sqrt{3}}+frac{sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}}

$$\frac{ \sqrt{ 5 } - \sqrt{ 3 } }{ \sqrt{ 5 } + \sqrt{ 3 } } + \frac{ \sqrt{ 5 } + \sqrt{ 3 } }{ \sqrt{ 5 } - \sqrt{ 3 } }$$

Evaluate

$$8$$

Solution Steps

Rationalize the denominator of \frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}} by multiplying numerator and denominator by \sqrt{5}-\sqrt{3}.

$$\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$$

Consider \left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{3}\right)^{2}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$$

Square \sqrt{5}. Square \sqrt{3}.

$$\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{5-3}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$$

Subtract 3 from 5 to get 2.

$$\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{2}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$$

Multiply \sqrt{5}-\sqrt{3} and \sqrt{5}-\sqrt{3} to get \left(\sqrt{5}-\sqrt{3}\right)^{2}.

$$\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{2}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{5}-\sqrt{3}\right)^{2}.

$$\frac{\left(\sqrt{5}\right)^{2}-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$$

The square of \sqrt{5} is 5.

$$\frac{5-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$$

To multiply \sqrt{5} and \sqrt{3}, multiply the numbers under the square root.

$$\frac{5-2\sqrt{15}+\left(\sqrt{3}\right)^{2}}{2}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$$

The square of \sqrt{3} is 3.

$$\frac{5-2\sqrt{15}+3}{2}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$$

Add 5 and 3 to get 8.

$$\frac{8-2\sqrt{15}}{2}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$$

Divide each term of 8-2\sqrt{15} by 2 to get 4-\sqrt{15}.

$$4-\sqrt{15}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$$

Rationalize the denominator of \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} by multiplying numerator and denominator by \sqrt{5}+\sqrt{3}.

$$4-\sqrt{15}+\frac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}$$

Consider \left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$4-\sqrt{15}+\frac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{3}\right)^{2}}$$

Square \sqrt{5}. Square \sqrt{3}.

$$4-\sqrt{15}+\frac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{5-3}$$

Subtract 3 from 5 to get 2.

$$4-\sqrt{15}+\frac{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{2}$$

Multiply \sqrt{5}+\sqrt{3} and \sqrt{5}+\sqrt{3} to get \left(\sqrt{5}+\sqrt{3}\right)^{2}.

$$4-\sqrt{15}+\frac{\left(\sqrt{5}+\sqrt{3}\right)^{2}}{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{5}+\sqrt{3}\right)^{2}.

$$4-\sqrt{15}+\frac{\left(\sqrt{5}\right)^{2}+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2}$$

The square of \sqrt{5} is 5.

$$4-\sqrt{15}+\frac{5+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2}$$

To multiply \sqrt{5} and \sqrt{3}, multiply the numbers under the square root.

$$4-\sqrt{15}+\frac{5+2\sqrt{15}+\left(\sqrt{3}\right)^{2}}{2}$$

The square of \sqrt{3} is 3.

$$4-\sqrt{15}+\frac{5+2\sqrt{15}+3}{2}$$

Add 5 and 3 to get 8.

$$4-\sqrt{15}+\frac{8+2\sqrt{15}}{2}$$

Divide each term of 8+2\sqrt{15} by 2 to get 4+\sqrt{15}.

$$4-\sqrt{15}+4+\sqrt{15}$$

Add 4 and 4 to get 8.

$$8-\sqrt{15}+\sqrt{15}$$

Combine -\sqrt{15} and \sqrt{15} to get 0.

$$8$$

Factor

$$2^{3}$$