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

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

Evaluate

$$14-2\sqrt{3}-3\sqrt{6}\approx 3.187429157$$

Solution Steps

Use the distributive property to multiply \sqrt{2} by \sqrt{3}-\sqrt{6}.

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

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

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

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}.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

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

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

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

The square of \sqrt{2} is 2.

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

Add 12 and 2 to get 14.

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

Combine \sqrt{6} and -4\sqrt{6} to get -3\sqrt{6}.

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