Solve (frac{sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}})^2+(frac{sqrt{5}-sqrt{3}}{sqrt{5}+sqrt{3}})^2

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

Evaluate

$$62$$

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}.

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

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}.

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

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

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

Subtract 3 from 5 to get 2.

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

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

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

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

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

The square of \sqrt{5} is 5.

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

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

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

The square of \sqrt{3} is 3.

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

Add 5 and 3 to get 8.

$$\left(\frac{8+2\sqrt{15}}{2}\right)^{2}+\left(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)^{2}$$

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

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4+\sqrt{15}\right)^{2}.

$$16+8\sqrt{15}+\left(\sqrt{15}\right)^{2}+\left(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)^{2}$$

The square of \sqrt{15} is 15.

$$16+8\sqrt{15}+15+\left(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)^{2}$$

Add 16 and 15 to get 31.

$$31+8\sqrt{15}+\left(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\right)^{2}$$

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

$$31+8\sqrt{15}+\left(\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}\right)^{2}$$

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}.

$$31+8\sqrt{15}+\left(\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}\right)^{2}-\left(\sqrt{3}\right)^{2}}\right)^{2}$$

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

$$31+8\sqrt{15}+\left(\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{5-3}\right)^{2}$$

Subtract 3 from 5 to get 2.

$$31+8\sqrt{15}+\left(\frac{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{2}\right)^{2}$$

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

$$31+8\sqrt{15}+\left(\frac{\left(\sqrt{5}-\sqrt{3}\right)^{2}}{2}\right)^{2}$$

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

$$31+8\sqrt{15}+\left(\frac{\left(\sqrt{5}\right)^{2}-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2}\right)^{2}$$

The square of \sqrt{5} is 5.

$$31+8\sqrt{15}+\left(\frac{5-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{2}\right)^{2}$$

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

$$31+8\sqrt{15}+\left(\frac{5-2\sqrt{15}+\left(\sqrt{3}\right)^{2}}{2}\right)^{2}$$

The square of \sqrt{3} is 3.

$$31+8\sqrt{15}+\left(\frac{5-2\sqrt{15}+3}{2}\right)^{2}$$

Add 5 and 3 to get 8.

$$31+8\sqrt{15}+\left(\frac{8-2\sqrt{15}}{2}\right)^{2}$$

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

$$31+8\sqrt{15}+\left(4-\sqrt{15}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-\sqrt{15}\right)^{2}.

$$31+8\sqrt{15}+16-8\sqrt{15}+\left(\sqrt{15}\right)^{2}$$

The square of \sqrt{15} is 15.

$$31+8\sqrt{15}+16-8\sqrt{15}+15$$

Add 16 and 15 to get 31.

$$31+8\sqrt{15}+31-8\sqrt{15}$$

Add 31 and 31 to get 62.

$$62+8\sqrt{15}-8\sqrt{15}$$

Combine 8\sqrt{15} and -8\sqrt{15} to get 0.

$$62$$

Factor

$$2\times 31$$