Solve frac{4-sqrt{2}}{4+sqrt{2}}-frac{4+sqrt{2}}{4-sqrt{2}}

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

Evaluate

$$-\frac{8\sqrt{2}}{7}\approx -1.616244071$$

Short Solution Steps

Rationalize the denominator of \frac{4-\sqrt{2}}{4+\sqrt{2}} by multiplying numerator and denominator by 4-\sqrt{2}.

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

Consider \left(4+\sqrt{2}\right)\left(4-\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square 4. Square \sqrt{2}.

$$\frac{\left(4-\sqrt{2}\right)\left(4-\sqrt{2}\right)}{16-2}-\frac{4+\sqrt{2}}{4-\sqrt{2}}$$

Subtract 2 from 16 to get 14.

$$\frac{\left(4-\sqrt{2}\right)\left(4-\sqrt{2}\right)}{14}-\frac{4+\sqrt{2}}{4-\sqrt{2}}$$

Multiply 4-\sqrt{2} and 4-\sqrt{2} to get \left(4-\sqrt{2}\right)^{2}.

$$\frac{\left(4-\sqrt{2}\right)^{2}}{14}-\frac{4+\sqrt{2}}{4-\sqrt{2}}$$

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

$$\frac{16-8\sqrt{2}+\left(\sqrt{2}\right)^{2}}{14}-\frac{4+\sqrt{2}}{4-\sqrt{2}}$$

The square of \sqrt{2} is 2.

$$\frac{16-8\sqrt{2}+2}{14}-\frac{4+\sqrt{2}}{4-\sqrt{2}}$$

Add 16 and 2 to get 18.

$$\frac{18-8\sqrt{2}}{14}-\frac{4+\sqrt{2}}{4-\sqrt{2}}$$

Rationalize the denominator of \frac{4+\sqrt{2}}{4-\sqrt{2}} by multiplying numerator and denominator by 4+\sqrt{2}.

$$\frac{18-8\sqrt{2}}{14}-\frac{\left(4+\sqrt{2}\right)\left(4+\sqrt{2}\right)}{\left(4-\sqrt{2}\right)\left(4+\sqrt{2}\right)}$$

Consider \left(4-\sqrt{2}\right)\left(4+\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{18-8\sqrt{2}}{14}-\frac{\left(4+\sqrt{2}\right)\left(4+\sqrt{2}\right)}{4^{2}-\left(\sqrt{2}\right)^{2}}$$

Square 4. Square \sqrt{2}.

$$\frac{18-8\sqrt{2}}{14}-\frac{\left(4+\sqrt{2}\right)\left(4+\sqrt{2}\right)}{16-2}$$

Subtract 2 from 16 to get 14.

$$\frac{18-8\sqrt{2}}{14}-\frac{\left(4+\sqrt{2}\right)\left(4+\sqrt{2}\right)}{14}$$

Multiply 4+\sqrt{2} and 4+\sqrt{2} to get \left(4+\sqrt{2}\right)^{2}.

$$\frac{18-8\sqrt{2}}{14}-\frac{\left(4+\sqrt{2}\right)^{2}}{14}$$

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

$$\frac{18-8\sqrt{2}}{14}-\frac{16+8\sqrt{2}+\left(\sqrt{2}\right)^{2}}{14}$$

The square of \sqrt{2} is 2.

$$\frac{18-8\sqrt{2}}{14}-\frac{16+8\sqrt{2}+2}{14}$$

Add 16 and 2 to get 18.

$$\frac{18-8\sqrt{2}}{14}-\frac{18+8\sqrt{2}}{14}$$

Since \frac{18-8\sqrt{2}}{14} and \frac{18+8\sqrt{2}}{14} have the same denominator, subtract them by subtracting their numerators.

$$\frac{18-8\sqrt{2}-\left(18+8\sqrt{2}\right)}{14}$$

Do the multiplications in 18-8\sqrt{2}-\left(18+8\sqrt{2}\right).

$$\frac{18-8\sqrt{2}-18-8\sqrt{2}}{14}$$

Do the calculations in 18-8\sqrt{2}-18-8\sqrt{2}.

$$\frac{-16\sqrt{2}}{14}$$

Divide -16\sqrt{2} by 14 to get -\frac{8}{7}\sqrt{2}.

$$-\frac{8}{7}\sqrt{2}$$