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

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

Evaluate

$$4\sqrt{6}+12\sqrt{2}-2\sqrt{3}\approx 23.304420104$$

Solution Steps

Subtract 1 from 3 to get 2.

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

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

$$4+8\sqrt{2}+4\left(\sqrt{2}\right)^{2}-\left(\sqrt{3}+1-2\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$4+8\sqrt{2}+4\times 2-\left(\sqrt{3}+1-2\sqrt{2}\right)^{2}$$

Multiply 4 and 2 to get 8.

$$4+8\sqrt{2}+8-\left(\sqrt{3}+1-2\sqrt{2}\right)^{2}$$

Add 4 and 8 to get 12.

$$12+8\sqrt{2}-\left(\sqrt{3}+1-2\sqrt{2}\right)^{2}$$

Square \sqrt{3}+1-2\sqrt{2}.

$$12+8\sqrt{2}-\left(-4\sqrt{2}\sqrt{3}+4\left(\sqrt{2}\right)^{2}+\left(\sqrt{3}\right)^{2}-4\sqrt{2}+2\sqrt{3}+1\right)$$

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

$$12+8\sqrt{2}-\left(-4\sqrt{6}+4\left(\sqrt{2}\right)^{2}+\left(\sqrt{3}\right)^{2}-4\sqrt{2}+2\sqrt{3}+1\right)$$

The square of \sqrt{2} is 2.

$$12+8\sqrt{2}-\left(-4\sqrt{6}+4\times 2+\left(\sqrt{3}\right)^{2}-4\sqrt{2}+2\sqrt{3}+1\right)$$

Multiply 4 and 2 to get 8.

$$12+8\sqrt{2}-\left(-4\sqrt{6}+8+\left(\sqrt{3}\right)^{2}-4\sqrt{2}+2\sqrt{3}+1\right)$$

The square of \sqrt{3} is 3.

$$12+8\sqrt{2}-\left(-4\sqrt{6}+8+3-4\sqrt{2}+2\sqrt{3}+1\right)$$

Add 8 and 3 to get 11.

$$12+8\sqrt{2}-\left(-4\sqrt{6}+11-4\sqrt{2}+2\sqrt{3}+1\right)$$

Add 11 and 1 to get 12.

$$12+8\sqrt{2}-\left(-4\sqrt{6}+12-4\sqrt{2}+2\sqrt{3}\right)$$

To find the opposite of -4\sqrt{6}+12-4\sqrt{2}+2\sqrt{3}, find the opposite of each term.

$$12+8\sqrt{2}+4\sqrt{6}-12+4\sqrt{2}-2\sqrt{3}$$

Subtract 12 from 12 to get 0.

$$8\sqrt{2}+4\sqrt{6}+4\sqrt{2}-2\sqrt{3}$$

Combine 8\sqrt{2} and 4\sqrt{2} to get 12\sqrt{2}.

$$12\sqrt{2}+4\sqrt{6}-2\sqrt{3}$$

Expand

$$4 \sqrt{6} + 12 \sqrt{2} - 2 \sqrt{3} = 23.304420104$$

Solution Steps

Subtract 1 from 3 to get 2.

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

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

$$4+8\sqrt{2}+4\left(\sqrt{2}\right)^{2}-\left(\sqrt{3}+1-2\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$4+8\sqrt{2}+4\times 2-\left(\sqrt{3}+1-2\sqrt{2}\right)^{2}$$

Multiply 4 and 2 to get 8.

$$4+8\sqrt{2}+8-\left(\sqrt{3}+1-2\sqrt{2}\right)^{2}$$

Add 4 and 8 to get 12.

$$12+8\sqrt{2}-\left(\sqrt{3}+1-2\sqrt{2}\right)^{2}$$

Square \sqrt{3}+1-2\sqrt{2}.

$$12+8\sqrt{2}-\left(-4\sqrt{2}\sqrt{3}+4\left(\sqrt{2}\right)^{2}+\left(\sqrt{3}\right)^{2}-4\sqrt{2}+2\sqrt{3}+1\right)$$

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

$$12+8\sqrt{2}-\left(-4\sqrt{6}+4\left(\sqrt{2}\right)^{2}+\left(\sqrt{3}\right)^{2}-4\sqrt{2}+2\sqrt{3}+1\right)$$

The square of \sqrt{2} is 2.

$$12+8\sqrt{2}-\left(-4\sqrt{6}+4\times 2+\left(\sqrt{3}\right)^{2}-4\sqrt{2}+2\sqrt{3}+1\right)$$

Multiply 4 and 2 to get 8.

$$12+8\sqrt{2}-\left(-4\sqrt{6}+8+\left(\sqrt{3}\right)^{2}-4\sqrt{2}+2\sqrt{3}+1\right)$$

The square of \sqrt{3} is 3.

$$12+8\sqrt{2}-\left(-4\sqrt{6}+8+3-4\sqrt{2}+2\sqrt{3}+1\right)$$

Add 8 and 3 to get 11.

$$12+8\sqrt{2}-\left(-4\sqrt{6}+11-4\sqrt{2}+2\sqrt{3}+1\right)$$

Add 11 and 1 to get 12.

$$12+8\sqrt{2}-\left(-4\sqrt{6}+12-4\sqrt{2}+2\sqrt{3}\right)$$

To find the opposite of -4\sqrt{6}+12-4\sqrt{2}+2\sqrt{3}, find the opposite of each term.

$$12+8\sqrt{2}+4\sqrt{6}-12+4\sqrt{2}-2\sqrt{3}$$

Subtract 12 from 12 to get 0.

$$8\sqrt{2}+4\sqrt{6}+4\sqrt{2}-2\sqrt{3}$$

Combine 8\sqrt{2} and 4\sqrt{2} to get 12\sqrt{2}.

$$12\sqrt{2}+4\sqrt{6}-2\sqrt{3}$$