Solve sqrt{128}-sqrt{98}+sqrt{18}=

$$\sqrt{ 128 } - \sqrt{ 98 } + \sqrt{ 18 } =$$

Evaluate

$$4\sqrt{2}\approx 5.656854249$$

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}.

$$8\sqrt{2}-\sqrt{98}+\sqrt{18}$$

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}.

$$8\sqrt{2}-7\sqrt{2}+\sqrt{18}$$

Combine 8\sqrt{2} and -7\sqrt{2} to get \sqrt{2}.

$$\sqrt{2}+\sqrt{18}$$

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}.

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

Combine \sqrt{2} and 3\sqrt{2} to get 4\sqrt{2}.

$$4\sqrt{2}$$