Solve (sqrt{6}+sqrt{2})*sqrt{3}

$$( \sqrt { 6 } + \sqrt { 2 } ) \times \sqrt { 3 }$$

Evaluate

$$\sqrt{6}+3\sqrt{2}\approx 6.69213043$$

Solution Steps

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

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

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

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

Multiply \sqrt{3} and \sqrt{3} to get 3.

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

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

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