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

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

Evaluate

$$\frac{3\sqrt{2}}{2}+1\approx 3.121320344$$

Short Solution Steps

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

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

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

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

Square 2. Square \sqrt{2}.

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

Subtract 2 from 4 to get 2.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2}{2}.

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

Since \frac{\left(\sqrt{2}+1\right)\left(2+\sqrt{2}\right)}{2} and \frac{2}{2} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in \left(\sqrt{2}+1\right)\left(2+\sqrt{2}\right)-2.

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

Do the calculations in 2\sqrt{2}+2+2+\sqrt{2}-2.

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