Solve 3sqrt{50}-2sqrt{18}+sqrt{98}

$$3 \sqrt{ 50 } -2 \sqrt{ 18 } + \sqrt{ 98 }$$

Evaluate

$$16\sqrt{2}\approx 22.627416998$$

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}.

$$3\times 5\sqrt{2}-2\sqrt{18}+\sqrt{98}$$

Multiply 3 and 5 to get 15.

$$15\sqrt{2}-2\sqrt{18}+\sqrt{98}$$

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}.

$$15\sqrt{2}-2\times 3\sqrt{2}+\sqrt{98}$$

Multiply -2 and 3 to get -6.

$$15\sqrt{2}-6\sqrt{2}+\sqrt{98}$$

Combine 15\sqrt{2} and -6\sqrt{2} to get 9\sqrt{2}.

$$9\sqrt{2}+\sqrt{98}$$

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}.

$$9\sqrt{2}+7\sqrt{2}$$

Combine 9\sqrt{2} and 7\sqrt{2} to get 16\sqrt{2}.

$$16\sqrt{2}$$