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

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

Evaluate

$$\frac{3\left(\sqrt{2}+4\right)}{7}\approx 2.320377241$$

Solution Steps

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

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

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}.

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

Square 3. Square \sqrt{2}.

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

Subtract 2 from 9 to get 7.

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

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

$$\frac{18-6\sqrt{2}+9\sqrt{2}-3\left(\sqrt{2}\right)^{2}}{7}$$

Combine -6\sqrt{2} and 9\sqrt{2} to get 3\sqrt{2}.

$$\frac{18+3\sqrt{2}-3\left(\sqrt{2}\right)^{2}}{7}$$

The square of \sqrt{2} is 2.

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

Multiply -3 and 2 to get -6.

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

Subtract 6 from 18 to get 12.

$$\frac{12+3\sqrt{2}}{7}$$

Factor

$$\frac{3 {(\sqrt{2} + 4)}}{7} = 2.3203772410170407$$