Solve sqrt{2}+sqrt{32}+sqrt{128}=

$$\sqrt { 2 } + \sqrt { 32 } + \sqrt { 128 } =$$

Evaluate

$$13\sqrt{2}\approx 18.384776311$$

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}.

$$\sqrt{2}+4\sqrt{2}+\sqrt{128}$$

Combine \sqrt{2} and 4\sqrt{2} to get 5\sqrt{2}.

$$5\sqrt{2}+\sqrt{128}$$

Factor 128=8^{2}\times 2. Rewrite the square root of the product \sqrt{8^{2}\times 2} as the product of square roots \sqrt{8^{2}}\sqrt{2}. Take the square root of 8^{2}.

$$5\sqrt{2}+8\sqrt{2}$$

Combine 5\sqrt{2} and 8\sqrt{2} to get 13\sqrt{2}.

$$13\sqrt{2}$$