Solve 7sqrt{12}+16sqrt{27}-9sqrt{300}

$$7 \sqrt { 12 } + 16 \sqrt { 27 } - 9 \sqrt { 300 }$$

Evaluate

$$-28\sqrt{3}\approx -48.497422612$$

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}.

$$7\times 2\sqrt{3}+16\sqrt{27}-9\sqrt{300}$$

Multiply 7 and 2 to get 14.

$$14\sqrt{3}+16\sqrt{27}-9\sqrt{300}$$

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}.

$$14\sqrt{3}+16\times 3\sqrt{3}-9\sqrt{300}$$

Multiply 16 and 3 to get 48.

$$14\sqrt{3}+48\sqrt{3}-9\sqrt{300}$$

Combine 14\sqrt{3} and 48\sqrt{3} to get 62\sqrt{3}.

$$62\sqrt{3}-9\sqrt{300}$$

Factor 300=10^{2}\times 3. Rewrite the square root of the product \sqrt{10^{2}\times 3} as the product of square roots \sqrt{10^{2}}\sqrt{3}. Take the square root of 10^{2}.

$$62\sqrt{3}-9\times 10\sqrt{3}$$

Multiply -9 and 10 to get -90.

$$62\sqrt{3}-90\sqrt{3}$$

Combine 62\sqrt{3} and -90\sqrt{3} to get -28\sqrt{3}.

$$-28\sqrt{3}$$