Solve {left(2sqrt{2}-3right)}^2-(2sqrt{2}+1)(2sqrt{2}-1)

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

Evaluate

$$10-12\sqrt{2}\approx -6.970562748$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{2}-3\right)^{2}.

$$4\left(\sqrt{2}\right)^{2}-12\sqrt{2}+9-\left(2\sqrt{2}+1\right)\left(2\sqrt{2}-1\right)$$

The square of \sqrt{2} is 2.

$$4\times 2-12\sqrt{2}+9-\left(2\sqrt{2}+1\right)\left(2\sqrt{2}-1\right)$$

Multiply 4 and 2 to get 8.

$$8-12\sqrt{2}+9-\left(2\sqrt{2}+1\right)\left(2\sqrt{2}-1\right)$$

Add 8 and 9 to get 17.

$$17-12\sqrt{2}-\left(2\sqrt{2}+1\right)\left(2\sqrt{2}-1\right)$$

Consider \left(2\sqrt{2}+1\right)\left(2\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}. Square 1.

$$17-12\sqrt{2}-\left(\left(2\sqrt{2}\right)^{2}-1\right)$$

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

$$17-12\sqrt{2}-\left(2^{2}\left(\sqrt{2}\right)^{2}-1\right)$$

Calculate 2 to the power of 2 and get 4.

$$17-12\sqrt{2}-\left(4\left(\sqrt{2}\right)^{2}-1\right)$$

The square of \sqrt{2} is 2.

$$17-12\sqrt{2}-\left(4\times 2-1\right)$$

Multiply 4 and 2 to get 8.

$$17-12\sqrt{2}-\left(8-1\right)$$

Subtract 1 from 8 to get 7.

$$17-12\sqrt{2}-7$$

Subtract 7 from 17 to get 10.

$$10-12\sqrt{2}$$