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

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

Evaluate

$$12\sqrt{2}\approx 16.970562748$$

Solution Steps

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

$$9+6\sqrt{2}+\left(\sqrt{2}\right)^{2}-\left(3-\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$9+6\sqrt{2}+2-\left(3-\sqrt{2}\right)^{2}$$

Add 9 and 2 to get 11.

$$11+6\sqrt{2}-\left(3-\sqrt{2}\right)^{2}$$

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

$$11+6\sqrt{2}-\left(9-6\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$11+6\sqrt{2}-\left(9-6\sqrt{2}+2\right)$$

Add 9 and 2 to get 11.

$$11+6\sqrt{2}-\left(11-6\sqrt{2}\right)$$

To find the opposite of 11-6\sqrt{2}, find the opposite of each term.

$$11+6\sqrt{2}-11+6\sqrt{2}$$

Subtract 11 from 11 to get 0.

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

Combine 6\sqrt{2} and 6\sqrt{2} to get 12\sqrt{2}.

$$12\sqrt{2}$$

Expand

$$12 \sqrt{2} = 16.970562748$$

Solution Steps

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

$$9+6\sqrt{2}+\left(\sqrt{2}\right)^{2}-\left(3-\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$9+6\sqrt{2}+2-\left(3-\sqrt{2}\right)^{2}$$

Add 9 and 2 to get 11.

$$11+6\sqrt{2}-\left(3-\sqrt{2}\right)^{2}$$

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

$$11+6\sqrt{2}-\left(9-6\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$11+6\sqrt{2}-\left(9-6\sqrt{2}+2\right)$$

Add 9 and 2 to get 11.

$$11+6\sqrt{2}-\left(11-6\sqrt{2}\right)$$

To find the opposite of 11-6\sqrt{2}, find the opposite of each term.

$$11+6\sqrt{2}-11+6\sqrt{2}$$

Subtract 11 from 11 to get 0.

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

Combine 6\sqrt{2} and 6\sqrt{2} to get 12\sqrt{2}.

$$12\sqrt{2}$$