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

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

Evaluate

$$4-\sqrt{2}\approx 2.585786438$$

Solution Steps

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

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

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}.

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

Square \sqrt{2}. Square 1.

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

Subtract 1 from 2 to get 1.

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

Anything divided by one gives itself.

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

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

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

The square of \sqrt{2} is 2.

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

Add 2 and 1 to get 3.

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

Add 1 and 3 to get 4.

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

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

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

Factor

$$4-\sqrt{2}$$