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

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

Evaluate

$$\frac{\sqrt{6}+6\sqrt{2}-4\sqrt{3}-2}{22}\approx 0.091207631$$

Solution Steps

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

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

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

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

The square of \sqrt{2} is 2.

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

Expand \left(-2\sqrt{6}\right)^{2}.

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

Calculate -2 to the power of 2 and get 4.

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

The square of \sqrt{6} is 6.

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

Multiply 4 and 6 to get 24.

$$\frac{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+2\sqrt{6}\right)}{2-24}$$

Subtract 24 from 2 to get -22.

$$\frac{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+2\sqrt{6}\right)}{-22}$$

Apply the distributive property by multiplying each term of \sqrt{2}-\sqrt{3} by each term of \sqrt{2}+2\sqrt{6}.

$$\frac{\left(\sqrt{2}\right)^{2}+2\sqrt{2}\sqrt{6}-\sqrt{3}\sqrt{2}-2\sqrt{3}\sqrt{6}}{-22}$$

The square of \sqrt{2} is 2.

$$\frac{2+2\sqrt{2}\sqrt{6}-\sqrt{3}\sqrt{2}-2\sqrt{3}\sqrt{6}}{-22}$$

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}.

$$\frac{2+2\sqrt{2}\sqrt{2}\sqrt{3}-\sqrt{3}\sqrt{2}-2\sqrt{3}\sqrt{6}}{-22}$$

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

$$\frac{2+2\times 2\sqrt{3}-\sqrt{3}\sqrt{2}-2\sqrt{3}\sqrt{6}}{-22}$$

Multiply 2 and 2 to get 4.

$$\frac{2+4\sqrt{3}-\sqrt{3}\sqrt{2}-2\sqrt{3}\sqrt{6}}{-22}$$

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

$$\frac{2+4\sqrt{3}-\sqrt{6}-2\sqrt{3}\sqrt{6}}{-22}$$

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

$$\frac{2+4\sqrt{3}-\sqrt{6}-2\sqrt{3}\sqrt{3}\sqrt{2}}{-22}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$\frac{2+4\sqrt{3}-\sqrt{6}-2\times 3\sqrt{2}}{-22}$$

Multiply -2 and 3 to get -6.

$$\frac{2+4\sqrt{3}-\sqrt{6}-6\sqrt{2}}{-22}$$

Multiply both numerator and denominator by -1.

$$\frac{-2-4\sqrt{3}+\sqrt{6}+6\sqrt{2}}{22}$$