Solve sqrt{8}-sqrt{32}+sqrt{128}

$$\sqrt { 8 } - \sqrt { 32 } + \sqrt { 128 }$$

Evaluate

$$6\sqrt{2}\approx 8.485281374$$

Solution Steps

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

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

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}.

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

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

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

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}.

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

Combine -2\sqrt{2} and 8\sqrt{2} to get 6\sqrt{2}.

$$6\sqrt{2}$$