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

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

Evaluate

$$13-9\sqrt{2}\approx 0.272077939$$

Solution Steps

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

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

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

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

Calculate 3 to the power of 2 and get 9.

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

Expand \left(2\sqrt{2}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(3-\sqrt{2}\right)\left(3-2\sqrt{2}\right)}{9-4\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\left(3-\sqrt{2}\right)\left(3-2\sqrt{2}\right)}{9-4\times 2}$$

Multiply 4 and 2 to get 8.

$$\frac{\left(3-\sqrt{2}\right)\left(3-2\sqrt{2}\right)}{9-8}$$

Subtract 8 from 9 to get 1.

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

Anything divided by one gives itself.

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

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

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

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

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

The square of \sqrt{2} is 2.

$$9-9\sqrt{2}+2\times 2$$

Multiply 2 and 2 to get 4.

$$9-9\sqrt{2}+4$$

Add 9 and 4 to get 13.

$$13-9\sqrt{2}$$