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

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

Evaluate

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

Solution Steps

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}.

$$7\sqrt{2}+\sqrt{242}-\sqrt{50}$$

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}.

$$7\sqrt{2}+11\sqrt{2}-\sqrt{50}$$

Combine 7\sqrt{2} and 11\sqrt{2} to get 18\sqrt{2}.

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

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}.

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

Combine 18\sqrt{2} and -5\sqrt{2} to get 13\sqrt{2}.

$$13\sqrt{2}$$