Solve frac{4+3sqrt{2}}{2+sqrt{3}}

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

Evaluate

$$6\sqrt{2}+8-3\sqrt{6}-4\sqrt{3}\approx 2.208608916$$

Solution Steps

Rationalize the denominator of \frac{4+3\sqrt{2}}{2+\sqrt{3}} by multiplying numerator and denominator by 2-\sqrt{3}.

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

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}.

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

Square 2. Square \sqrt{3}.

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

Subtract 3 from 4 to get 1.

$$\frac{\left(4+3\sqrt{2}\right)\left(2-\sqrt{3}\right)}{1}$$

Anything divided by one gives itself.

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

Apply the distributive property by multiplying each term of 4+3\sqrt{2} by each term of 2-\sqrt{3}.

$$8-4\sqrt{3}+6\sqrt{2}-3\sqrt{3}\sqrt{2}$$

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

$$8-4\sqrt{3}+6\sqrt{2}-3\sqrt{6}$$