Solve sqrt{200}-sqrt{128}+sqrt{68}

$$\sqrt { 200 } - \sqrt { 128 } + \sqrt { 68 }$$

Evaluate

$$2\left(\sqrt{2}+\sqrt{17}\right)\approx 11.074638376$$

Solution Steps

Factor 200=10^{2}\times 2. Rewrite the square root of the product \sqrt{10^{2}\times 2} as the product of square roots \sqrt{10^{2}}\sqrt{2}. Take the square root of 10^{2}.

$$10\sqrt{2}-\sqrt{128}+\sqrt{68}$$

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}.

$$10\sqrt{2}-8\sqrt{2}+\sqrt{68}$$

Combine 10\sqrt{2} and -8\sqrt{2} to get 2\sqrt{2}.

$$2\sqrt{2}+\sqrt{68}$$

Factor 68=2^{2}\times 17. Rewrite the square root of the product \sqrt{2^{2}\times 17} as the product of square roots \sqrt{2^{2}}\sqrt{17}. Take the square root of 2^{2}.

$$2\sqrt{2}+2\sqrt{17}$$