Solve sqrt[3]{2}-sqrt{32}+sqrt{50}+sqrt{98}

$$\sqrt[ 3 ]{ 2 } - \sqrt{ 32 } + \sqrt{ 50 } + \sqrt{ 98 }$$

Evaluate

$$\sqrt[3]{2}+8\sqrt{2}\approx 12.573629549$$

Solution Steps

Factor 32=4^{2}\times 2. Rewrite the square root of the product \sqrt{4^{2}\times 2} as the product of square roots \sqrt{4^{2}}\sqrt{2}. Take the square root of 4^{2}.

$$\sqrt[3]{2}-4\sqrt{2}+\sqrt{50}+\sqrt{98}$$

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}.

$$\sqrt[3]{2}-4\sqrt{2}+5\sqrt{2}+\sqrt{98}$$

Combine -4\sqrt{2} and 5\sqrt{2} to get \sqrt{2}.

$$\sqrt[3]{2}+\sqrt{2}+\sqrt{98}$$

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}.

$$\sqrt[3]{2}+\sqrt{2}+7\sqrt{2}$$

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

$$\sqrt[3]{2}+8\sqrt{2}$$