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

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

Evaluate

$$21-6\sqrt{6}\approx 6.303061543$$

Solution Steps

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

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

The square of \sqrt{2} is 2.

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

Multiply 9 and 2 to get 18.

$$18-6\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}$$

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

$$18-6\sqrt{6}+\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

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

Add 18 and 3 to get 21.

$$21-6\sqrt{6}$$

Expand

$$21-6\sqrt{6}$$

Solution Steps

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

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

The square of \sqrt{2} is 2.

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

Multiply 9 and 2 to get 18.

$$18-6\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}$$

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

$$18-6\sqrt{6}+\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

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

Add 18 and 3 to get 21.

$$21-6\sqrt{6}$$