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

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

Evaluate

$$\sqrt{3}-2\approx -0.267949192$$

Solution Steps

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

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

Consider \left(\sqrt{2}+\sqrt{6}\right)\left(\sqrt{2}-\sqrt{6}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{2}. Square \sqrt{6}.

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

Subtract 6 from 2 to get -4.

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

Multiply \sqrt{2}-\sqrt{6} and \sqrt{2}-\sqrt{6} to get \left(\sqrt{2}-\sqrt{6}\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{2}-\sqrt{6}\right)^{2}.

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

The square of \sqrt{2} is 2.

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

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$\frac{2-2\sqrt{2}\sqrt{2}\sqrt{3}+\left(\sqrt{6}\right)^{2}}{-4}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{2-2\times 2\sqrt{3}+\left(\sqrt{6}\right)^{2}}{-4}$$

Multiply -2 and 2 to get -4.

$$\frac{2-4\sqrt{3}+\left(\sqrt{6}\right)^{2}}{-4}$$

The square of \sqrt{6} is 6.

$$\frac{2-4\sqrt{3}+6}{-4}$$

Add 2 and 6 to get 8.

$$\frac{8-4\sqrt{3}}{-4}$$

Divide each term of 8-4\sqrt{3} by -4 to get -2+\sqrt{3}.

$$-2+\sqrt{3}$$