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

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

Evaluate

$$-3\sqrt{10}-47\approx -56.486832981$$

Solution Steps

Use the distributive property to multiply 2\sqrt{5}+5\sqrt{2} by \sqrt{5}-5\sqrt{2} and combine like terms.

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

The square of \sqrt{5} is 5.

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

Multiply 2 and 5 to get 10.

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

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

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

The square of \sqrt{2} is 2.

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

Multiply -25 and 2 to get -50.

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

Subtract 50 from 10 to get -40.

$$-40-5\sqrt{10}-\left(\sqrt{5}-\sqrt{2}\right)^{2}$$

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

$$-40-5\sqrt{10}-\left(\left(\sqrt{5}\right)^{2}-2\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{5} is 5.

$$-40-5\sqrt{10}-\left(5-2\sqrt{5}\sqrt{2}+\left(\sqrt{2}\right)^{2}\right)$$

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

$$-40-5\sqrt{10}-\left(5-2\sqrt{10}+\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$-40-5\sqrt{10}-\left(5-2\sqrt{10}+2\right)$$

Add 5 and 2 to get 7.

$$-40-5\sqrt{10}-\left(7-2\sqrt{10}\right)$$

To find the opposite of 7-2\sqrt{10}, find the opposite of each term.

$$-40-5\sqrt{10}-7+2\sqrt{10}$$

Subtract 7 from -40 to get -47.

$$-47-5\sqrt{10}+2\sqrt{10}$$

Combine -5\sqrt{10} and 2\sqrt{10} to get -3\sqrt{10}.

$$-47-3\sqrt{10}$$