Solve (sqrt{2}-2sqrt{3})^2+4sqrt{1/3}*3sqrt{2}

$$( \sqrt { 2 } - 2 \sqrt { 3 } ) ^ { 2 } + 4 \sqrt { \frac { 1 } { 3 } } \times 3 \sqrt { 2 }$$

Evaluate

$$14$$

Solution Steps

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

$$\left(\sqrt{2}\right)^{2}-4\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}+3\times 4\sqrt{\frac{1}{3}}\sqrt{2}$$

The square of \sqrt{2} is 2.

$$2-4\sqrt{2}\sqrt{3}+4\left(\sqrt{3}\right)^{2}+3\times 4\sqrt{\frac{1}{3}}\sqrt{2}$$

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

$$2-4\sqrt{6}+4\left(\sqrt{3}\right)^{2}+3\times 4\sqrt{\frac{1}{3}}\sqrt{2}$$

The square of \sqrt{3} is 3.

$$2-4\sqrt{6}+4\times 3+3\times 4\sqrt{\frac{1}{3}}\sqrt{2}$$

Multiply 4 and 3 to get 12.

$$2-4\sqrt{6}+12+3\times 4\sqrt{\frac{1}{3}}\sqrt{2}$$

Add 2 and 12 to get 14.

$$14-4\sqrt{6}+3\times 4\sqrt{\frac{1}{3}}\sqrt{2}$$

Multiply 3 and 4 to get 12.

$$14-4\sqrt{6}+12\sqrt{\frac{1}{3}}\sqrt{2}$$

Rewrite the square root of the division \sqrt{\frac{1}{3}} as the division of square roots \frac{\sqrt{1}}{\sqrt{3}}.

$$14-4\sqrt{6}+12\times \frac{\sqrt{1}}{\sqrt{3}}\sqrt{2}$$

Calculate the square root of 1 and get 1.

$$14-4\sqrt{6}+12\times \frac{1}{\sqrt{3}}\sqrt{2}$$

Rationalize the denominator of \frac{1}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$14-4\sqrt{6}+12\times \frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}\sqrt{2}$$

The square of \sqrt{3} is 3.

$$14-4\sqrt{6}+12\times \frac{\sqrt{3}}{3}\sqrt{2}$$

Cancel out 3, the greatest common factor in 12 and 3.

$$14-4\sqrt{6}+4\sqrt{3}\sqrt{2}$$

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

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

Combine -4\sqrt{6} and 4\sqrt{6} to get 0.

$$14$$

Factor

$$2\times 7$$