Solve (frac{sqrt{2}+1}{sqrt{2}-1})

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

Evaluate

$$2\sqrt{2}+3\approx 5.828427125$$

Solution Steps

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

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

Consider \left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{2}. Square 1.

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

Subtract 1 from 2 to get 1.

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

Anything divided by one gives itself.

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

Multiply \sqrt{2}+1 and \sqrt{2}+1 to get \left(\sqrt{2}+1\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{2}+1\right)^{2}.

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

The square of \sqrt{2} is 2.

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

Add 2 and 1 to get 3.

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