Solve frac{2sqrt{3}-sqrt{6}}{sqrt{2}-1}

$$\frac { 2 \sqrt { 3 } - \sqrt { 6 } } { \sqrt { 2 } - 1 }$$

Evaluate

$$\sqrt{6}\approx 2.449489743$$

Solution Steps

Rationalize the denominator of \frac{2\sqrt{3}-\sqrt{6}}{\sqrt{2}-1} by multiplying numerator and denominator by \sqrt{2}+1.

$$\frac{\left(2\sqrt{3}-\sqrt{6}\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}$$

Consider \left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(2\sqrt{3}-\sqrt{6}\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}\right)^{2}-1^{2}}$$

Square \sqrt{2}. Square 1.

$$\frac{\left(2\sqrt{3}-\sqrt{6}\right)\left(\sqrt{2}+1\right)}{2-1}$$

Subtract 1 from 2 to get 1.

$$\frac{\left(2\sqrt{3}-\sqrt{6}\right)\left(\sqrt{2}+1\right)}{1}$$

Anything divided by one gives itself.

$$\left(2\sqrt{3}-\sqrt{6}\right)\left(\sqrt{2}+1\right)$$

Apply the distributive property by multiplying each term of 2\sqrt{3}-\sqrt{6} by each term of \sqrt{2}+1.

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

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

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

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}.

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

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

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

Combine 2\sqrt{3} and -2\sqrt{3} to get 0.

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

Combine 2\sqrt{6} and -\sqrt{6} to get \sqrt{6}.

$$\sqrt{6}$$