Solve frac{sqrt{2}}{sqrt{6}-sqrt{2}}-frac{sqrt{3}}{sqrt{6}+sqrt{2}}

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

Evaluate

$$\frac{\sqrt{6}+2\sqrt{3}+2-3\sqrt{2}}{4}\approx 0.917737668$$

Short Solution Steps

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

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

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

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

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

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

Subtract 2 from 6 to get 4.

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

Rationalize the denominator of \frac{\sqrt{3}}{\sqrt{6}+\sqrt{2}} by multiplying numerator and denominator by \sqrt{6}-\sqrt{2}.

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

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

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

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

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

Subtract 2 from 6 to get 4.

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

Since \frac{\sqrt{2}\left(\sqrt{6}+\sqrt{2}\right)}{4} and \frac{\sqrt{3}\left(\sqrt{6}-\sqrt{2}\right)}{4} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \sqrt{2}\left(\sqrt{6}+\sqrt{2}\right)-\sqrt{3}\left(\sqrt{6}-\sqrt{2}\right).

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