Solve sqrt{32}-sqrt{72}+3sqrt{98}=

$$\sqrt { 32 } - \sqrt { 72 } + 3 \sqrt { 98 } =$$

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}.

$$4\sqrt{2}-\sqrt{72}+3\sqrt{98}$$

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}.

$$4\sqrt{2}-6\sqrt{2}+3\sqrt{98}$$

Combine 4\sqrt{2} and -6\sqrt{2} to get -2\sqrt{2}.

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

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}.

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

Multiply 3 and 7 to get 21.

$$-2\sqrt{2}+21\sqrt{2}$$

Combine -2\sqrt{2} and 21\sqrt{2} to get 19\sqrt{2}.

$$19\sqrt{2}$$