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

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

Evaluate

$$\frac{6\sqrt{2}+11}{7}\approx 2.783611625$$

Solution Steps

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

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

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}.

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

Square 3. Square \sqrt{2}.

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

Subtract 2 from 9 to get 7.

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

Multiply 3+\sqrt{2} and 3+\sqrt{2} to get \left(3+\sqrt{2}\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+\sqrt{2}\right)^{2}.

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

The square of \sqrt{2} is 2.

$$\frac{9+6\sqrt{2}+2}{7}$$

Add 9 and 2 to get 11.

$$\frac{11+6\sqrt{2}}{7}$$