Solve sqrt{6}*(5)(2sqrt{5}+5sqrt{2})^2-(2sqrt{5}-5sqrt{2})^2

$$\sqrt { 6 } \cdot ( 5 ) ( 2 \sqrt { 5 } + 5 \sqrt { 2 } ) ^ { 2 } - ( 2 \sqrt { 5 } - 5 \sqrt { 2 } ) ^ { 2 }$$

Evaluate

$$20\sqrt{10}+200\sqrt{15}+350\sqrt{6}-70\approx 1625.163632419$$

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}.

$$\sqrt{6}\times 5\left(4\left(\sqrt{5}\right)^{2}+20\sqrt{5}\sqrt{2}+25\left(\sqrt{2}\right)^{2}\right)-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

The square of \sqrt{5} is 5.

$$\sqrt{6}\times 5\left(4\times 5+20\sqrt{5}\sqrt{2}+25\left(\sqrt{2}\right)^{2}\right)-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

Multiply 4 and 5 to get 20.

$$\sqrt{6}\times 5\left(20+20\sqrt{5}\sqrt{2}+25\left(\sqrt{2}\right)^{2}\right)-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

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

$$\sqrt{6}\times 5\left(20+20\sqrt{10}+25\left(\sqrt{2}\right)^{2}\right)-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$\sqrt{6}\times 5\left(20+20\sqrt{10}+25\times 2\right)-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

Multiply 25 and 2 to get 50.

$$\sqrt{6}\times 5\left(20+20\sqrt{10}+50\right)-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

Add 20 and 50 to get 70.

$$\sqrt{6}\times 5\left(70+20\sqrt{10}\right)-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

Use the distributive property to multiply \sqrt{6}\times 5 by 70+20\sqrt{10}.

$$70\sqrt{6}\times 5+100\sqrt{6}\sqrt{10}-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

Multiply 70 and 5 to get 350.

$$350\sqrt{6}+100\sqrt{6}\sqrt{10}-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

To multiply \sqrt{6} and \sqrt{10}, multiply the numbers under the square root.

$$350\sqrt{6}+100\sqrt{60}-\left(2\sqrt{5}-5\sqrt{2}\right)^{2}$$

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

$$350\sqrt{6}+100\sqrt{60}-\left(4\left(\sqrt{5}\right)^{2}-20\sqrt{5}\sqrt{2}+25\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{5} is 5.

$$350\sqrt{6}+100\sqrt{60}-\left(4\times 5-20\sqrt{5}\sqrt{2}+25\left(\sqrt{2}\right)^{2}\right)$$

Multiply 4 and 5 to get 20.

$$350\sqrt{6}+100\sqrt{60}-\left(20-20\sqrt{5}\sqrt{2}+25\left(\sqrt{2}\right)^{2}\right)$$

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

$$350\sqrt{6}+100\sqrt{60}-\left(20-20\sqrt{10}+25\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$350\sqrt{6}+100\sqrt{60}-\left(20-20\sqrt{10}+25\times 2\right)$$

Multiply 25 and 2 to get 50.

$$350\sqrt{6}+100\sqrt{60}-\left(20-20\sqrt{10}+50\right)$$

Add 20 and 50 to get 70.

$$350\sqrt{6}+100\sqrt{60}-\left(70-20\sqrt{10}\right)$$

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

$$350\sqrt{6}+100\sqrt{60}-70+20\sqrt{10}$$

Factor 60=2^{2}\times 15. Rewrite the square root of the product \sqrt{2^{2}\times 15} as the product of square roots \sqrt{2^{2}}\sqrt{15}. Take the square root of 2^{2}.

$$350\sqrt{6}+100\times 2\sqrt{15}-70+20\sqrt{10}$$

Multiply 100 and 2 to get 200.

$$350\sqrt{6}+200\sqrt{15}-70+20\sqrt{10}$$