Solve {left(sqrt{6}-2sqrt{3}right)}^2

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

Evaluate

$$18-12\sqrt{2}\approx 1.029437252$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{6}-2\sqrt{3}\right)^{2}.

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

The square of \sqrt{6} is 6.

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

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

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

Multiply \sqrt{3} and \sqrt{3} to get 3.

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

Multiply -4 and 3 to get -12.

$$6-12\sqrt{2}+4\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$6-12\sqrt{2}+4\times 3$$

Multiply 4 and 3 to get 12.

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

Add 6 and 12 to get 18.

$$18-12\sqrt{2}$$

Expand

$$18-12\sqrt{2}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{6}-2\sqrt{3}\right)^{2}.

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

The square of \sqrt{6} is 6.

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

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

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

Multiply \sqrt{3} and \sqrt{3} to get 3.

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

Multiply -4 and 3 to get -12.

$$6-12\sqrt{2}+4\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$6-12\sqrt{2}+4\times 3$$

Multiply 4 and 3 to get 12.

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

Add 6 and 12 to get 18.

$$18-12\sqrt{2}$$