Solve {left(2sqrt{5}+5sqrt{2}right)}^2-{left(2sqrt{5}-5sqrt{2}right)}^2

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

Evaluate

$$40\sqrt{10}\approx 126.491106407$$

Solution Steps

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

$$4\left(\sqrt{5}\right)^{2}+20\sqrt{5}\sqrt{2}+25\left(\sqrt{2}\right)^{2}-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

The square of \sqrt{5} is 5.

$$4\times 5+20\sqrt{5}\sqrt{2}+25\left(\sqrt{2}\right)^{2}-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

Multiply 4 and 5 to get 20.

$$20+20\sqrt{5}\sqrt{2}+25\left(\sqrt{2}\right)^{2}-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

To multiply \sqrt{5} and \sqrt{2}, multiply the numbers under the square root.

$$20+20\sqrt{10}+25\left(\sqrt{2}\right)^{2}-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$20+20\sqrt{10}+25\times 2-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

Multiply 25 and 2 to get 50.

$$20+20\sqrt{10}+50-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

Add 20 and 50 to get 70.

$$70+20\sqrt{10}-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

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

$$70+20\sqrt{10}-\left(4\left(\sqrt{5}\right)^{2}-20\sqrt{5}\sqrt{2}+25\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{5} is 5.

$$70+20\sqrt{10}-\left(4\times 5-20\sqrt{5}\sqrt{2}+25\left(\sqrt{2}\right)^{2}\right)$$

Multiply 4 and 5 to get 20.

$$70+20\sqrt{10}-\left(20-20\sqrt{5}\sqrt{2}+25\left(\sqrt{2}\right)^{2}\right)$$

To multiply \sqrt{5} and \sqrt{2}, multiply the numbers under the square root.

$$70+20\sqrt{10}-\left(20-20\sqrt{10}+25\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$70+20\sqrt{10}-\left(20-20\sqrt{10}+25\times 2\right)$$

Multiply 25 and 2 to get 50.

$$70+20\sqrt{10}-\left(20-20\sqrt{10}+50\right)$$

Add 20 and 50 to get 70.

$$70+20\sqrt{10}-\left(70-20\sqrt{10}\right)$$

To find the opposite of 70-20\sqrt{10}, find the opposite of each term.

$$70+20\sqrt{10}-70+20\sqrt{10}$$

Subtract 70 from 70 to get 0.

$$20\sqrt{10}+20\sqrt{10}$$

Combine 20\sqrt{10} and 20\sqrt{10} to get 40\sqrt{10}.

$$40\sqrt{10}$$

Expand

$$40 \sqrt{10} = 126.491106407$$

Solution Steps

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

$$4\left(\sqrt{5}\right)^{2}+20\sqrt{5}\sqrt{2}+25\left(\sqrt{2}\right)^{2}-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

The square of \sqrt{5} is 5.

$$4\times 5+20\sqrt{5}\sqrt{2}+25\left(\sqrt{2}\right)^{2}-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

Multiply 4 and 5 to get 20.

$$20+20\sqrt{5}\sqrt{2}+25\left(\sqrt{2}\right)^{2}-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

To multiply \sqrt{5} and \sqrt{2}, multiply the numbers under the square root.

$$20+20\sqrt{10}+25\left(\sqrt{2}\right)^{2}-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$20+20\sqrt{10}+25\times 2-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

Multiply 25 and 2 to get 50.

$$20+20\sqrt{10}+50-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

Add 20 and 50 to get 70.

$$70+20\sqrt{10}-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

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

$$70+20\sqrt{10}-\left(4\left(\sqrt{5}\right)^{2}-20\sqrt{5}\sqrt{2}+25\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{5} is 5.

$$70+20\sqrt{10}-\left(4\times 5-20\sqrt{5}\sqrt{2}+25\left(\sqrt{2}\right)^{2}\right)$$

Multiply 4 and 5 to get 20.

$$70+20\sqrt{10}-\left(20-20\sqrt{5}\sqrt{2}+25\left(\sqrt{2}\right)^{2}\right)$$

To multiply \sqrt{5} and \sqrt{2}, multiply the numbers under the square root.

$$70+20\sqrt{10}-\left(20-20\sqrt{10}+25\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$70+20\sqrt{10}-\left(20-20\sqrt{10}+25\times 2\right)$$

Multiply 25 and 2 to get 50.

$$70+20\sqrt{10}-\left(20-20\sqrt{10}+50\right)$$

Add 20 and 50 to get 70.

$$70+20\sqrt{10}-\left(70-20\sqrt{10}\right)$$

To find the opposite of 70-20\sqrt{10}, find the opposite of each term.

$$70+20\sqrt{10}-70+20\sqrt{10}$$

Subtract 70 from 70 to get 0.

$$20\sqrt{10}+20\sqrt{10}$$

Combine 20\sqrt{10} and 20\sqrt{10} to get 40\sqrt{10}.

$$40\sqrt{10}$$