Solve sqrt{32}-3*sqrt{2}+2*sqrt{18}

$$\sqrt { 32 } - 3 \cdot \sqrt { 2 } + 2 \cdot \sqrt { 18 }$$

Evaluate

$$7\sqrt{2}\approx 9.899494937$$

Solution Steps

Factor 32=4^{2}\times 2. Rewrite the square root of the product \sqrt{4^{2}\times 2} as the product of square roots \sqrt{4^{2}}\sqrt{2}. Take the square root of 4^{2}.

$$4\sqrt{2}-3\sqrt{2}+2\sqrt{18}$$

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

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

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}.

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

Multiply 2 and 3 to get 6.

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

Combine \sqrt{2} and 6\sqrt{2} to get 7\sqrt{2}.

$$7\sqrt{2}$$