Solve 1/sqrt{2+sqrt{6}}

$$\frac { 1 } { \sqrt { 2 } + \sqrt { 6 } }$$

Evaluate

$$\frac{\sqrt{6}-\sqrt{2}}{4}\approx 0.258819045$$

Solution Steps

Rationalize the denominator of \frac{1}{\sqrt{2}+\sqrt{6}} by multiplying numerator and denominator by \sqrt{2}-\sqrt{6}.

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

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}.

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

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

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

Subtract 6 from 2 to get -4.

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

Multiply both numerator and denominator by -1.

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