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

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

Evaluate

$$\sqrt{2}-2\approx -0.585786438$$

Solution Steps

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

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

The square of \sqrt{2} is 2.

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

Add 2 and 1 to get 3.

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

To find the opposite of 3-2\sqrt{2}, find the opposite of each term.

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

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

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

Subtract 1 from -3 to get -4.

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

Add -4 and 2 to get -2.

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