Solve =(2/sqrt{7+sqrt{5}}-2/sqrt{7-sqrt{5}})^2

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

Evaluate

$$20$$

Solution Steps

Rationalize the denominator of \frac{2}{\sqrt{7}+\sqrt{5}} by multiplying numerator and denominator by \sqrt{7}-\sqrt{5}.

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

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}.

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

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

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

Subtract 5 from 7 to get 2.

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

Cancel out 2 and 2.

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

Rationalize the denominator of \frac{2}{\sqrt{7}-\sqrt{5}} by multiplying numerator and denominator by \sqrt{7}+\sqrt{5}.

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

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}.

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

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

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

Subtract 5 from 7 to get 2.

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

Cancel out 2 and 2.

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

To find the opposite of \sqrt{7}+\sqrt{5}, find the opposite of each term.

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

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

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

Combine -\sqrt{5} and -\sqrt{5} to get -2\sqrt{5}.

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

Expand \left(-2\sqrt{5}\right)^{2}.

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

Calculate -2 to the power of 2 and get 4.

$$4\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$4\times 5$$

Multiply 4 and 5 to get 20.

$$20$$

Factor

$$2^{2}\times 5$$