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

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

Evaluate

$$4\left(\sqrt{2}+1\right)\approx 9.656854249$$

Solution Steps

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

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

Calculate -\sqrt{2} to the power of 2 and get \left(\sqrt{2}\right)^{2}.

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

Multiply -2 and -1 to get 2.

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

Use the distributive property to multiply 2 by \left(\sqrt{2}\right)^{2}+2\sqrt{2}+1.

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

The square of \sqrt{2} is 2.

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

Multiply 2 and 2 to get 4.

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

Add 4 and 2 to get 6.

$$6+4\sqrt{2}-2$$

Subtract 2 from 6 to get 4.

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

Expand

$$4 \sqrt{2} + 4 = 9.656854249$$

Solution Steps

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

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

Calculate -\sqrt{2} to the power of 2 and get \left(\sqrt{2}\right)^{2}.

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

Multiply -2 and -1 to get 2.

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

Use the distributive property to multiply 2 by \left(\sqrt{2}\right)^{2}+2\sqrt{2}+1.

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

The square of \sqrt{2} is 2.

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

Multiply 2 and 2 to get 4.

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

Add 4 and 2 to get 6.

$$6+4\sqrt{2}-2$$

Subtract 2 from 6 to get 4.

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