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

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

Evaluate

$$\frac{\sqrt{2}+4}{7}\approx 0.77345908$$

Solution Steps

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

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

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}.

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

Square 3. Square \sqrt{2}.

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

Subtract 2 from 9 to get 7.

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

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

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

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

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

The square of \sqrt{2} is 2.

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

Subtract 2 from 6 to get 4.

$$\frac{4+\sqrt{2}}{7}$$

Factor

$$\frac{\sqrt{2} + 4}{7} = 0.7734590803390136$$