Solve 8sqrt{27}-9sqrt{12}-sqrt{75}quad

$$8 \sqrt { 27 } - 9 \sqrt { 12 } - \sqrt { 75 } \quad$$

Evaluate

$$\sqrt{3}\approx 1.732050808$$

Solution Steps

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

$$8\times 3\sqrt{3}-9\sqrt{12}-\sqrt{75}$$

Multiply 8 and 3 to get 24.

$$24\sqrt{3}-9\sqrt{12}-\sqrt{75}$$

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}.

$$24\sqrt{3}-9\times 2\sqrt{3}-\sqrt{75}$$

Multiply -9 and 2 to get -18.

$$24\sqrt{3}-18\sqrt{3}-\sqrt{75}$$

Combine 24\sqrt{3} and -18\sqrt{3} to get 6\sqrt{3}.

$$6\sqrt{3}-\sqrt{75}$$

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}.

$$6\sqrt{3}-5\sqrt{3}$$

Combine 6\sqrt{3} and -5\sqrt{3} to get \sqrt{3}.

$$\sqrt{3}$$