Solve frac{sqrt{2}-1}{3-2sqrt{2}}=

$$\frac{ \sqrt{ 2 } -1 }{ 3-2 \sqrt{ 2 } } =$$

Evaluate

$$\sqrt{2}+1\approx 2.414213562$$

Solution Steps

Rationalize the denominator of \frac{\sqrt{2}-1}{3-2\sqrt{2}} by multiplying numerator and denominator by 3+2\sqrt{2}.

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

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}.

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

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)}{9-\left(-2\sqrt{2}\right)^{2}}$$

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

$$\frac{\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)}{9-\left(-2\right)^{2}\left(\sqrt{2}\right)^{2}}$$

Calculate -2 to the power of 2 and get 4.

$$\frac{\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)}{9-4\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)}{9-4\times 2}$$

Multiply 4 and 2 to get 8.

$$\frac{\left(\sqrt{2}-1\right)\left(3+2\sqrt{2}\right)}{9-8}$$

Subtract 8 from 9 to get 1.

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

Anything divided by one gives itself.

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

Apply the distributive property by multiplying each term of \sqrt{2}-1 by each term of 3+2\sqrt{2}.

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

The square of \sqrt{2} is 2.

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

Multiply 2 and 2 to get 4.

$$3\sqrt{2}+4-3-2\sqrt{2}$$

Subtract 3 from 4 to get 1.

$$3\sqrt{2}+1-2\sqrt{2}$$

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

$$\sqrt{2}+1$$