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

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

Evaluate

$$6-2\sqrt{6}\approx 1.101020514$$

Solution Steps

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

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

The square of \sqrt{2} is 2.

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

Multiply 3 and 2 to get 6.

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

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

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

Combine -3\sqrt{6} and \sqrt{6} to get -2\sqrt{6}.

$$6-2\sqrt{6}$$