Solve 6sqrt{12}+3sqrt{27}-7sqrt{3}+sqrt{18}

$$6 \sqrt { 12 } + 3 \sqrt { 27 } - 7 \sqrt { 3 } + \sqrt { 18 }$$

Evaluate

$$\sqrt{3}\left(\sqrt{6}+14\right)\approx 28.491351993$$

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}.

$$6\times 2\sqrt{3}+3\sqrt{27}-7\sqrt{3}+\sqrt{18}$$

Multiply 6 and 2 to get 12.

$$12\sqrt{3}+3\sqrt{27}-7\sqrt{3}+\sqrt{18}$$

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}.

$$12\sqrt{3}+3\times 3\sqrt{3}-7\sqrt{3}+\sqrt{18}$$

Multiply 3 and 3 to get 9.

$$12\sqrt{3}+9\sqrt{3}-7\sqrt{3}+\sqrt{18}$$

Combine 12\sqrt{3} and 9\sqrt{3} to get 21\sqrt{3}.

$$21\sqrt{3}-7\sqrt{3}+\sqrt{18}$$

Combine 21\sqrt{3} and -7\sqrt{3} to get 14\sqrt{3}.

$$14\sqrt{3}+\sqrt{18}$$

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}.

$$14\sqrt{3}+3\sqrt{2}$$