Solve 3sqrt{32}+sqrt{98}

$$3 \sqrt { 32 } + \sqrt { 98 }$$

Evaluate

$$19\sqrt{2}\approx 26.870057685$$

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}.

$$3\times 4\sqrt{2}+\sqrt{98}$$

Multiply 3 and 4 to get 12.

$$12\sqrt{2}+\sqrt{98}$$

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}.

$$12\sqrt{2}+7\sqrt{2}$$

Combine 12\sqrt{2} and 7\sqrt{2} to get 19\sqrt{2}.

$$19\sqrt{2}$$