Solve sqrt{128}+sqrt{50}-sqrt{98}-sqrt{242}

$$\sqrt { 128 } + \sqrt { 50 } - \sqrt { 98 } - \sqrt { 242 }$$

Evaluate

$$-5\sqrt{2}\approx -7.071067812$$

Solution Steps

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}.

$$8\sqrt{2}+\sqrt{50}-\sqrt{98}-\sqrt{242}$$

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}.

$$8\sqrt{2}+5\sqrt{2}-\sqrt{98}-\sqrt{242}$$

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

$$13\sqrt{2}-\sqrt{98}-\sqrt{242}$$

Factor 98=7^{2}\times 2. Rewrite the square root of the product \sqrt{7^{2}\times 2} as the product of square roots \sqrt{7^{2}}\sqrt{2}. Take the square root of 7^{2}.

$$13\sqrt{2}-7\sqrt{2}-\sqrt{242}$$

Combine 13\sqrt{2} and -7\sqrt{2} to get 6\sqrt{2}.

$$6\sqrt{2}-\sqrt{242}$$

Factor 242=11^{2}\times 2. Rewrite the square root of the product \sqrt{11^{2}\times 2} as the product of square roots \sqrt{11^{2}}\sqrt{2}. Take the square root of 11^{2}.

$$6\sqrt{2}-11\sqrt{2}$$

Combine 6\sqrt{2} and -11\sqrt{2} to get -5\sqrt{2}.

$$-5\sqrt{2}$$