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

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

Evaluate

$$\sqrt{3}\approx 1.732050808$$

Solution Steps

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

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

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

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

Square \sqrt{3}. Square \sqrt{2}.

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

Subtract 2 from 3 to get 1.

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

Anything divided by one gives itself.

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

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

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

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}.

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

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

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

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

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

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}.

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

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

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

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

$$\sqrt{3}$$