Solve (3sqrt{5}-2)^2-(2sqrt{2})^2

$$( 3 \sqrt { 5 } - 2 ) ^ { 2 } - ( 2 \sqrt { 2 } ) ^ { 2 }$$

Evaluate

$$41-12\sqrt{5}\approx 14.16718427$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3\sqrt{5}-2\right)^{2}.

$$9\left(\sqrt{5}\right)^{2}-12\sqrt{5}+4-\left(2\sqrt{2}\right)^{2}$$

The square of \sqrt{5} is 5.

$$9\times 5-12\sqrt{5}+4-\left(2\sqrt{2}\right)^{2}$$

Multiply 9 and 5 to get 45.

$$45-12\sqrt{5}+4-\left(2\sqrt{2}\right)^{2}$$

Add 45 and 4 to get 49.

$$49-12\sqrt{5}-\left(2\sqrt{2}\right)^{2}$$

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

$$49-12\sqrt{5}-2^{2}\left(\sqrt{2}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$49-12\sqrt{5}-4\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$49-12\sqrt{5}-4\times 2$$

Multiply 4 and 2 to get 8.

$$49-12\sqrt{5}-8$$

Subtract 8 from 49 to get 41.

$$41-12\sqrt{5}$$

Expand

$$41-12\sqrt{5}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3\sqrt{5}-2\right)^{2}.

$$9\left(\sqrt{5}\right)^{2}-12\sqrt{5}+4-\left(2\sqrt{2}\right)^{2}$$

The square of \sqrt{5} is 5.

$$9\times 5-12\sqrt{5}+4-\left(2\sqrt{2}\right)^{2}$$

Multiply 9 and 5 to get 45.

$$45-12\sqrt{5}+4-\left(2\sqrt{2}\right)^{2}$$

Add 45 and 4 to get 49.

$$49-12\sqrt{5}-\left(2\sqrt{2}\right)^{2}$$

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

$$49-12\sqrt{5}-2^{2}\left(\sqrt{2}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$49-12\sqrt{5}-4\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$49-12\sqrt{5}-4\times 2$$

Multiply 4 and 2 to get 8.

$$49-12\sqrt{5}-8$$

Subtract 8 from 49 to get 41.

$$41-12\sqrt{5}$$