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

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

Evaluate

$$14-4\sqrt{6}\approx 4.202041029$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$12-4\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}$$

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

$$12-4\sqrt{6}+\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$12-4\sqrt{6}+2$$

Add 12 and 2 to get 14.

$$14-4\sqrt{6}$$

Expand

$$14-4\sqrt{6}$$

Solution Steps

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

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$12-4\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}$$

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

$$12-4\sqrt{6}+\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$12-4\sqrt{6}+2$$

Add 12 and 2 to get 14.

$$14-4\sqrt{6}$$