Solve sqrt{8}+sqrt{50}-sqrt{128}+sqrt{18}=

$$\sqrt { 8 } + \sqrt { 50 } - \sqrt { 128 } + \sqrt { 18 } =$$

Evaluate

$$2\sqrt{2}\approx 2.828427125$$

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}.

$$2\sqrt{2}+\sqrt{50}-\sqrt{128}+\sqrt{18}$$

Factor 50=5^{2}\times 2. Rewrite the square root of the product \sqrt{5^{2}\times 2} as the product of square roots \sqrt{5^{2}}\sqrt{2}. Take the square root of 5^{2}.

$$2\sqrt{2}+5\sqrt{2}-\sqrt{128}+\sqrt{18}$$

Combine 2\sqrt{2} and 5\sqrt{2} to get 7\sqrt{2}.

$$7\sqrt{2}-\sqrt{128}+\sqrt{18}$$

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}.

$$7\sqrt{2}-8\sqrt{2}+\sqrt{18}$$

Combine 7\sqrt{2} and -8\sqrt{2} to get -\sqrt{2}.

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

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}.

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

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

$$2\sqrt{2}$$