Solve (3sqrt{3}-5sqrt{2}+4sqrt{72}-2sqrt{18})

$$( 3 \sqrt { 3 } - 5 \sqrt { 2 } + 4 \sqrt { 72 } - 2 \sqrt { 18 } )$$

Evaluate

$$3\sqrt{3}+13\sqrt{2}\approx 23.580928734$$

Solution Steps

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}.

$$3\sqrt{3}-5\sqrt{2}+4\times 6\sqrt{2}-2\sqrt{18}$$

Multiply 4 and 6 to get 24.

$$3\sqrt{3}-5\sqrt{2}+24\sqrt{2}-2\sqrt{18}$$

Combine -5\sqrt{2} and 24\sqrt{2} to get 19\sqrt{2}.

$$3\sqrt{3}+19\sqrt{2}-2\sqrt{18}$$

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}.

$$3\sqrt{3}+19\sqrt{2}-2\times 3\sqrt{2}$$

Multiply -2 and 3 to get -6.

$$3\sqrt{3}+19\sqrt{2}-6\sqrt{2}$$

Combine 19\sqrt{2} and -6\sqrt{2} to get 13\sqrt{2}.

$$3\sqrt{3}+13\sqrt{2}$$