Solve sqrt{6}div(sqrt{6}-sqrt{2})

$$\sqrt { 6 } \div ( \sqrt { 6 } - \sqrt { 2 } )$$

Evaluate

$$\frac{\sqrt{3}+3}{2}\approx 2.366025404$$

Solution Steps

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

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

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}.

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

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

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

Subtract 2 from 6 to get 4.

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

Use the distributive property to multiply \sqrt{6} by \sqrt{6}+\sqrt{2}.

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

The square of \sqrt{6} is 6.

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

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}.

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

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

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