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

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

Evaluate

$$24-12\sqrt{3}\approx 3.215390309$$

Solution Steps

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

$$\left(\sqrt{6}\right)^{2}-6\sqrt{6}\sqrt{2}+9\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{6} is 6.

$$6-6\sqrt{6}\sqrt{2}+9\left(\sqrt{2}\right)^{2}$$

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

$$6-6\sqrt{2}\sqrt{3}\sqrt{2}+9\left(\sqrt{2}\right)^{2}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$6-6\times 2\sqrt{3}+9\left(\sqrt{2}\right)^{2}$$

Multiply -6 and 2 to get -12.

$$6-12\sqrt{3}+9\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$6-12\sqrt{3}+9\times 2$$

Multiply 9 and 2 to get 18.

$$6-12\sqrt{3}+18$$

Add 6 and 18 to get 24.

$$24-12\sqrt{3}$$

Expand

$$24-12\sqrt{3}$$

Solution Steps

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

$$\left(\sqrt{6}\right)^{2}-6\sqrt{6}\sqrt{2}+9\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{6} is 6.

$$6-6\sqrt{6}\sqrt{2}+9\left(\sqrt{2}\right)^{2}$$

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

$$6-6\sqrt{2}\sqrt{3}\sqrt{2}+9\left(\sqrt{2}\right)^{2}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$6-6\times 2\sqrt{3}+9\left(\sqrt{2}\right)^{2}$$

Multiply -6 and 2 to get -12.

$$6-12\sqrt{3}+9\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$6-12\sqrt{3}+9\times 2$$

Multiply 9 and 2 to get 18.

$$6-12\sqrt{3}+18$$

Add 6 and 18 to get 24.

$$24-12\sqrt{3}$$