Solve sqrt{6}*sqrt{3}+2sqrt{18}

$$\sqrt { 6 } \times \sqrt { 3 } + 2 \sqrt { 18 }$$

Evaluate

$$9\sqrt{2}\approx 12.727922061$$

Solution Steps

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}.

$$\sqrt{3}\sqrt{2}\sqrt{3}+2\sqrt{18}$$

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

$$3\sqrt{2}+2\sqrt{18}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

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

Multiply 2 and 3 to get 6.

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

Combine 3\sqrt{2} and 6\sqrt{2} to get 9\sqrt{2}.

$$9\sqrt{2}$$