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

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

Evaluate

$$\sqrt{6}+2\sqrt{2}+3\sqrt{3}+6\approx 16.47406929$$

Solution Steps

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

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

Consider \left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square 2. Square \sqrt{3}.

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

Subtract 3 from 4 to get 1.

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

Anything divided by one gives itself.

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

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

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

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

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