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

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

Evaluate

$$42-24\sqrt{3}\approx 0.430780618$$

Solution Steps

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

$$4\left(\sqrt{6}\right)^{2}-12\sqrt{6}\sqrt{2}+9\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{6} is 6.

$$4\times 6-12\sqrt{6}\sqrt{2}+9\left(\sqrt{2}\right)^{2}$$

Multiply 4 and 6 to get 24.

$$24-12\sqrt{6}\sqrt{2}+9\left(\sqrt{2}\right)^{2}$$

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}.

$$24-12\sqrt{2}\sqrt{3}\sqrt{2}+9\left(\sqrt{2}\right)^{2}$$

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

$$24-12\times 2\sqrt{3}+9\left(\sqrt{2}\right)^{2}$$

Multiply -12 and 2 to get -24.

$$24-24\sqrt{3}+9\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$24-24\sqrt{3}+9\times 2$$

Multiply 9 and 2 to get 18.

$$24-24\sqrt{3}+18$$

Add 24 and 18 to get 42.

$$42-24\sqrt{3}$$

Expand

$$42-24\sqrt{3}$$

Solution Steps

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

$$4\left(\sqrt{6}\right)^{2}-12\sqrt{6}\sqrt{2}+9\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{6} is 6.

$$4\times 6-12\sqrt{6}\sqrt{2}+9\left(\sqrt{2}\right)^{2}$$

Multiply 4 and 6 to get 24.

$$24-12\sqrt{6}\sqrt{2}+9\left(\sqrt{2}\right)^{2}$$

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}.

$$24-12\sqrt{2}\sqrt{3}\sqrt{2}+9\left(\sqrt{2}\right)^{2}$$

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

$$24-12\times 2\sqrt{3}+9\left(\sqrt{2}\right)^{2}$$

Multiply -12 and 2 to get -24.

$$24-24\sqrt{3}+9\left(\sqrt{2}\right)^{2}$$

The square of \sqrt{2} is 2.

$$24-24\sqrt{3}+9\times 2$$

Multiply 9 and 2 to get 18.

$$24-24\sqrt{3}+18$$

Add 24 and 18 to get 42.

$$42-24\sqrt{3}$$