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

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

Evaluate

$$\frac{-\sqrt{6}-2\sqrt{3}}{3}\approx -1.971197119$$

Solution Steps

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

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

Consider \left(\sqrt{3}-\sqrt{6}\right)\left(\sqrt{3}+\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}.

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

Square \sqrt{3}. Square \sqrt{6}.

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

Subtract 6 from 3 to get -3.

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

Use the distributive property to multiply \sqrt{2} by \sqrt{3}+\sqrt{6}.

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

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

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

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}.

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

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

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

Multiply both numerator and denominator by -1.

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