Solve (10sqrt{48}-6sqrt{27}+4sqrt{12})+sqrt{3}

$$( 10 \sqrt { 48 } - 6 \sqrt { 27 } + 4 \sqrt { 12 } ) + \sqrt { 3 }$$

Evaluate

$$31\sqrt{3}\approx 53.693575035$$

Solution Steps

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

$$10\times 4\sqrt{3}-6\sqrt{27}+4\sqrt{12}+\sqrt{3}$$

Multiply 10 and 4 to get 40.

$$40\sqrt{3}-6\sqrt{27}+4\sqrt{12}+\sqrt{3}$$

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}.

$$40\sqrt{3}-6\times 3\sqrt{3}+4\sqrt{12}+\sqrt{3}$$

Multiply -6 and 3 to get -18.

$$40\sqrt{3}-18\sqrt{3}+4\sqrt{12}+\sqrt{3}$$

Combine 40\sqrt{3} and -18\sqrt{3} to get 22\sqrt{3}.

$$22\sqrt{3}+4\sqrt{12}+\sqrt{3}$$

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}.

$$22\sqrt{3}+4\times 2\sqrt{3}+\sqrt{3}$$

Multiply 4 and 2 to get 8.

$$22\sqrt{3}+8\sqrt{3}+\sqrt{3}$$

Combine 22\sqrt{3} and 8\sqrt{3} to get 30\sqrt{3}.

$$30\sqrt{3}+\sqrt{3}$$

Combine 30\sqrt{3} and \sqrt{3} to get 31\sqrt{3}.

$$31\sqrt{3}$$