Solve 6sqrt{3}+4sqrt{27}+4sqrt{12}

$$6 \sqrt{ 3 } +4 \sqrt{ 27 } +4 \sqrt{ 12 }$$

Evaluate

$$26\sqrt{3}\approx 45.033320997$$

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}.

$$6\sqrt{3}+4\times 3\sqrt{3}+4\sqrt{12}$$

Multiply 4 and 3 to get 12.

$$6\sqrt{3}+12\sqrt{3}+4\sqrt{12}$$

Combine 6\sqrt{3} and 12\sqrt{3} to get 18\sqrt{3}.

$$18\sqrt{3}+4\sqrt{12}$$

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}.

$$18\sqrt{3}+4\times 2\sqrt{3}$$

Multiply 4 and 2 to get 8.

$$18\sqrt{3}+8\sqrt{3}$$

Combine 18\sqrt{3} and 8\sqrt{3} to get 26\sqrt{3}.

$$26\sqrt{3}$$