Solve sqrt{3}*(sqrt{12}+sqrt{2})

$$\sqrt { 3 } \cdot ( \sqrt { 12 } + \sqrt { 2 } )$$

Evaluate

$$\sqrt{6}+6\approx 8.449489743$$

Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 2 and 3 to get 6.

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

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

$$6+\sqrt{6}$$