Solve 6sqrt{8}+3sqrt{18}-10sqrt{32}

$$6 \sqrt { 8 } + 3 \sqrt { 18 } - 10 \sqrt { 32 }$$

Evaluate

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

Solution Steps

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$6\times 2\sqrt{2}+3\sqrt{18}-10\sqrt{32}$$

Multiply 6 and 2 to get 12.

$$12\sqrt{2}+3\sqrt{18}-10\sqrt{32}$$

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}.

$$12\sqrt{2}+3\times 3\sqrt{2}-10\sqrt{32}$$

Multiply 3 and 3 to get 9.

$$12\sqrt{2}+9\sqrt{2}-10\sqrt{32}$$

Combine 12\sqrt{2} and 9\sqrt{2} to get 21\sqrt{2}.

$$21\sqrt{2}-10\sqrt{32}$$

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}.

$$21\sqrt{2}-10\times 4\sqrt{2}$$

Multiply -10 and 4 to get -40.

$$21\sqrt{2}-40\sqrt{2}$$

Combine 21\sqrt{2} and -40\sqrt{2} to get -19\sqrt{2}.

$$-19\sqrt{2}$$