Solve sqrt{98}-3sqrt{72}+sqrt{50}

$$\sqrt { 98 } - 3 \sqrt { 72 } + \sqrt { 50 }$$

Evaluate

$$-6\sqrt{2}\approx -8.485281374$$

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}.

$$7\sqrt{2}-3\sqrt{72}+\sqrt{50}$$

Factor 72=6^{2}\times 2. Rewrite the square root of the product \sqrt{6^{2}\times 2} as the product of square roots \sqrt{6^{2}}\sqrt{2}. Take the square root of 6^{2}.

$$7\sqrt{2}-3\times 6\sqrt{2}+\sqrt{50}$$

Multiply -3 and 6 to get -18.

$$7\sqrt{2}-18\sqrt{2}+\sqrt{50}$$

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

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

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}.

$$-11\sqrt{2}+5\sqrt{2}$$

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

$$-6\sqrt{2}$$