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

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

Evaluate

$$\frac{3\sqrt{3}+18-\sqrt{6}-6\sqrt{2}}{7}\approx 1.751625901$$

Solution Steps

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

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

Consider \left(3+\sqrt{2}\right)\left(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}.

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

Square 3. Square \sqrt{2}.

$$\frac{\left(\sqrt{3}+6\right)\left(3-\sqrt{2}\right)}{9-2}$$

Subtract 2 from 9 to get 7.

$$\frac{\left(\sqrt{3}+6\right)\left(3-\sqrt{2}\right)}{7}$$

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

$$\frac{3\sqrt{3}-\sqrt{3}\sqrt{2}+18-6\sqrt{2}}{7}$$

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

$$\frac{3\sqrt{3}-\sqrt{6}+18-6\sqrt{2}}{7}$$