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

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

Evaluate

$$62-20\sqrt{6}\approx 13.010205144$$

Solution Steps

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

$$25\left(\sqrt{2}\right)^{2}-20\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{2} is 2.

$$25\times 2-20\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}$$

Multiply 25 and 2 to get 50.

$$50-20\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}$$

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

$$50-20\sqrt{6}+4\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$50-20\sqrt{6}+4\times 3$$

Multiply 4 and 3 to get 12.

$$50-20\sqrt{6}+12$$

Add 50 and 12 to get 62.

$$62-20\sqrt{6}$$

Expand

$$62-20\sqrt{6}$$

Solution Steps

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

$$25\left(\sqrt{2}\right)^{2}-20\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{2} is 2.

$$25\times 2-20\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}$$

Multiply 25 and 2 to get 50.

$$50-20\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}$$

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

$$50-20\sqrt{6}+4\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$50-20\sqrt{6}+4\times 3$$

Multiply 4 and 3 to get 12.

$$50-20\sqrt{6}+12$$

Add 50 and 12 to get 62.

$$62-20\sqrt{6}$$