Solve 4sqrt{3}-sqrt{2}left(sqrt{2}+2sqrt{3}right)+{left(2-sqrt{3}right)}^2

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

Evaluate

$$5-2\sqrt{6}\approx 0.101020514$$

Solution Steps

Use the distributive property to multiply \sqrt{2} by \sqrt{2}+2\sqrt{3}.

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

The square of \sqrt{2} is 2.

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

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

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

To find the opposite of 2+2\sqrt{6}, find the opposite of each term.

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

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

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

The square of \sqrt{3} is 3.

$$4\sqrt{3}-2-2\sqrt{6}+4-4\sqrt{3}+3$$

Add 4 and 3 to get 7.

$$4\sqrt{3}-2-2\sqrt{6}+7-4\sqrt{3}$$

Add -2 and 7 to get 5.

$$4\sqrt{3}+5-2\sqrt{6}-4\sqrt{3}$$

Combine 4\sqrt{3} and -4\sqrt{3} to get 0.

$$5-2\sqrt{6}$$