Solve 2/2(sqrt{2-1)}

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

Evaluate

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

Solution Steps

Use the distributive property to multiply 2 by \sqrt{2}-1.

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

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

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

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{2} is 2.

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

Multiply 4 and 2 to get 8.

$$\frac{2\left(2\sqrt{2}+2\right)}{8-2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{2\left(2\sqrt{2}+2\right)}{8-4}$$

Subtract 4 from 8 to get 4.

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

Divide 2\left(2\sqrt{2}+2\right) by 4 to get \frac{1}{2}\left(2\sqrt{2}+2\right).

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

Use the distributive property to multiply \frac{1}{2} by 2\sqrt{2}+2.

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

Cancel out 2 and 2.

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

Cancel out 2 and 2.

$$\sqrt{2}+1$$