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

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

Evaluate

$$2\sqrt{3}+3\sqrt{2}\approx 7.706742302$$

Solution Steps

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

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

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}.

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

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

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

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}.

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

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

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