Solve (sqrt{2}-1)^2+2sqrt{8}-(sqrt{5}-2)(sqrt{5}+2)

$$( \sqrt { 2 } - 1 ) ^ { 2 } + 2 \sqrt { 8 } - ( \sqrt { 5 } - 2 ) ( \sqrt { 5 } + 2 )$$

Evaluate

$$2\sqrt{2}+2\approx 4.828427125$$

Solution Steps

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

$$\left(\sqrt{2}\right)^{2}-2\sqrt{2}+1+2\sqrt{8}-\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)$$

The square of \sqrt{2} is 2.

$$2-2\sqrt{2}+1+2\sqrt{8}-\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)$$

Add 2 and 1 to get 3.

$$3-2\sqrt{2}+2\sqrt{8}-\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$3-2\sqrt{2}+2\times 2\sqrt{2}-\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)$$

Multiply 2 and 2 to get 4.

$$3-2\sqrt{2}+4\sqrt{2}-\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)$$

Combine -2\sqrt{2} and 4\sqrt{2} to get 2\sqrt{2}.

$$3+2\sqrt{2}-\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)$$

Consider \left(\sqrt{5}-2\right)\left(\sqrt{5}+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}. Square 2.

$$3+2\sqrt{2}-\left(\left(\sqrt{5}\right)^{2}-4\right)$$

The square of \sqrt{5} is 5.

$$3+2\sqrt{2}-\left(5-4\right)$$

Subtract 4 from 5 to get 1.

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

Subtract 1 from 3 to get 2.

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