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

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

Evaluate

$$4\sqrt{3}+12\sqrt{2}+10\approx 33.898765979$$

Solution Steps

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

$$9+12\sqrt{2}+4\left(\sqrt{2}\right)^{2}-\left(2-\sqrt{3}\right)^{2}$$

The square of \sqrt{2} is 2.

$$9+12\sqrt{2}+4\times 2-\left(2-\sqrt{3}\right)^{2}$$

Multiply 4 and 2 to get 8.

$$9+12\sqrt{2}+8-\left(2-\sqrt{3}\right)^{2}$$

Add 9 and 8 to get 17.

$$17+12\sqrt{2}-\left(2-\sqrt{3}\right)^{2}$$

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

$$17+12\sqrt{2}-\left(4-4\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$17+12\sqrt{2}-\left(4-4\sqrt{3}+3\right)$$

Add 4 and 3 to get 7.

$$17+12\sqrt{2}-\left(7-4\sqrt{3}\right)$$

To find the opposite of 7-4\sqrt{3}, find the opposite of each term.

$$17+12\sqrt{2}-7+4\sqrt{3}$$

Subtract 7 from 17 to get 10.

$$10+12\sqrt{2}+4\sqrt{3}$$

Expand

$$4 \sqrt{3} + 12 \sqrt{2} + 10 = 33.898765979$$

Solution Steps

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

$$9+12\sqrt{2}+4\left(\sqrt{2}\right)^{2}-\left(2-\sqrt{3}\right)^{2}$$

The square of \sqrt{2} is 2.

$$9+12\sqrt{2}+4\times 2-\left(2-\sqrt{3}\right)^{2}$$

Multiply 4 and 2 to get 8.

$$9+12\sqrt{2}+8-\left(2-\sqrt{3}\right)^{2}$$

Add 9 and 8 to get 17.

$$17+12\sqrt{2}-\left(2-\sqrt{3}\right)^{2}$$

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

$$17+12\sqrt{2}-\left(4-4\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$17+12\sqrt{2}-\left(4-4\sqrt{3}+3\right)$$

Add 4 and 3 to get 7.

$$17+12\sqrt{2}-\left(7-4\sqrt{3}\right)$$

To find the opposite of 7-4\sqrt{3}, find the opposite of each term.

$$17+12\sqrt{2}-7+4\sqrt{3}$$

Subtract 7 from 17 to get 10.

$$10+12\sqrt{2}+4\sqrt{3}$$