Solve sqrt{50}-sqrt{300}-sqrt{98}+sqrt{363}=

$$\sqrt { 50 } - \sqrt { 300 } - \sqrt { 98 } + \sqrt { 363 } =$$

Evaluate

$$\sqrt{3}-2\sqrt{2}\approx -1.096376317$$

Solution Steps

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}.

$$5\sqrt{2}-\sqrt{300}-\sqrt{98}+\sqrt{363}$$

Factor 300=10^{2}\times 3. Rewrite the square root of the product \sqrt{10^{2}\times 3} as the product of square roots \sqrt{10^{2}}\sqrt{3}. Take the square root of 10^{2}.

$$5\sqrt{2}-10\sqrt{3}-\sqrt{98}+\sqrt{363}$$

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}.

$$5\sqrt{2}-10\sqrt{3}-7\sqrt{2}+\sqrt{363}$$

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

$$-2\sqrt{2}-10\sqrt{3}+\sqrt{363}$$

Factor 363=11^{2}\times 3. Rewrite the square root of the product \sqrt{11^{2}\times 3} as the product of square roots \sqrt{11^{2}}\sqrt{3}. Take the square root of 11^{2}.

$$-2\sqrt{2}-10\sqrt{3}+11\sqrt{3}$$

Combine -10\sqrt{3} and 11\sqrt{3} to get \sqrt{3}.

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