Solve sqrt{12}+sqrt{6}*sqrt{2}=

$$\sqrt { 12 } + \sqrt { 6 } \cdot \sqrt { 2 } =$$

Evaluate

$$4\sqrt{3}\approx 6.92820323$$

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}.

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

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}.

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

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

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

Combine 2\sqrt{3} and 2\sqrt{3} to get 4\sqrt{3}.

$$4\sqrt{3}$$