Solve 40sqrt{3}-3sqrt{12}-20sqrt{12}+40sqrt{6}-2sqrt{3}

$$40 \sqrt { 3 } - 3 \sqrt { 12 } - 20 \sqrt { 12 } + 40 \sqrt { 6 } - 2 \sqrt { 3 }$$

Evaluate

$$40\sqrt{6}-8\sqrt{3}\approx 84.123183251$$

Solution Steps

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}.

$$40\sqrt{3}-3\times 2\sqrt{3}-20\sqrt{12}+40\sqrt{6}-2\sqrt{3}$$

Multiply -3 and 2 to get -6.

$$40\sqrt{3}-6\sqrt{3}-20\sqrt{12}+40\sqrt{6}-2\sqrt{3}$$

Combine 40\sqrt{3} and -6\sqrt{3} to get 34\sqrt{3}.

$$34\sqrt{3}-20\sqrt{12}+40\sqrt{6}-2\sqrt{3}$$

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}.

$$34\sqrt{3}-20\times 2\sqrt{3}+40\sqrt{6}-2\sqrt{3}$$

Multiply -20 and 2 to get -40.

$$34\sqrt{3}-40\sqrt{3}+40\sqrt{6}-2\sqrt{3}$$

Combine 34\sqrt{3} and -40\sqrt{3} to get -6\sqrt{3}.

$$-6\sqrt{3}+40\sqrt{6}-2\sqrt{3}$$

Combine -6\sqrt{3} and -2\sqrt{3} to get -8\sqrt{3}.

$$-8\sqrt{3}+40\sqrt{6}$$