Solve 2/sqrt{6+2}-2/sqrt{6-2}

$$\frac{ 2 }{ \sqrt{ 6 } +2 } - \frac{ 2 }{ \sqrt{ 6 } -2 }$$

Evaluate

$$-4$$

Solution Steps

Rationalize the denominator of \frac{2}{\sqrt{6}+2} by multiplying numerator and denominator by \sqrt{6}-2.

$$\frac{2\left(\sqrt{6}-2\right)}{\left(\sqrt{6}+2\right)\left(\sqrt{6}-2\right)}-\frac{2}{\sqrt{6}-2}$$

Consider \left(\sqrt{6}+2\right)\left(\sqrt{6}-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}.

$$\frac{2\left(\sqrt{6}-2\right)}{\left(\sqrt{6}\right)^{2}-2^{2}}-\frac{2}{\sqrt{6}-2}$$

Square \sqrt{6}. Square 2.

$$\frac{2\left(\sqrt{6}-2\right)}{6-4}-\frac{2}{\sqrt{6}-2}$$

Subtract 4 from 6 to get 2.

$$\frac{2\left(\sqrt{6}-2\right)}{2}-\frac{2}{\sqrt{6}-2}$$

Cancel out 2 and 2.

$$\sqrt{6}-2-\frac{2}{\sqrt{6}-2}$$

Rationalize the denominator of \frac{2}{\sqrt{6}-2} by multiplying numerator and denominator by \sqrt{6}+2.

$$\sqrt{6}-2-\frac{2\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}$$

Consider \left(\sqrt{6}-2\right)\left(\sqrt{6}+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}.

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

Square \sqrt{6}. Square 2.

$$\sqrt{6}-2-\frac{2\left(\sqrt{6}+2\right)}{6-4}$$

Subtract 4 from 6 to get 2.

$$\sqrt{6}-2-\frac{2\left(\sqrt{6}+2\right)}{2}$$

Cancel out 2 and 2.

$$\sqrt{6}-2-\left(\sqrt{6}+2\right)$$

To find the opposite of \sqrt{6}+2, find the opposite of each term.

$$\sqrt{6}-2-\sqrt{6}-2$$

Combine \sqrt{6} and -\sqrt{6} to get 0.

$$-2-2$$

Subtract 2 from -2 to get -4.

$$-4$$

Factor

$$-4$$