Solve sqrt{32}-2sqrt{12}-sqrt{75}+3sqrt{72}

$$\sqrt{ 32 } -2 \sqrt{ 12 } - \sqrt{ 75 } +3 \sqrt{ 72 }$$

Evaluate

$$22\sqrt{2}-9\sqrt{3}\approx 15.524241104$$

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}.

$$4\sqrt{2}-2\sqrt{12}-\sqrt{75}+3\sqrt{72}$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$4\sqrt{2}-2\times 2\sqrt{3}-\sqrt{75}+3\sqrt{72}$$

Multiply -2 and 2 to get -4.

$$4\sqrt{2}-4\sqrt{3}-\sqrt{75}+3\sqrt{72}$$

Factor 75=5^{2}\times 3. Rewrite the square root of the product \sqrt{5^{2}\times 3} as the product of square roots \sqrt{5^{2}}\sqrt{3}. Take the square root of 5^{2}.

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

Combine -4\sqrt{3} and -5\sqrt{3} to get -9\sqrt{3}.

$$4\sqrt{2}-9\sqrt{3}+3\sqrt{72}$$

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}.

$$4\sqrt{2}-9\sqrt{3}+3\times 6\sqrt{2}$$

Multiply 3 and 6 to get 18.

$$4\sqrt{2}-9\sqrt{3}+18\sqrt{2}$$

Combine 4\sqrt{2} and 18\sqrt{2} to get 22\sqrt{2}.

$$22\sqrt{2}-9\sqrt{3}$$