Solve {r}{2sqrt{12}-6sqrt{3}}{+3sqrt{48}}

$$\left. \begin{array} { r } { 2 \sqrt { 12 } - 6 \sqrt { 3 } } \\ { + 3 \sqrt { 48 } } \end{array} \right.$$

Evaluate

$$10\sqrt{3}\approx 17.320508076$$

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}.

$$2\times 2\sqrt{3}-6\sqrt{3}+3\sqrt{48}$$

Multiply 2 and 2 to get 4.

$$4\sqrt{3}-6\sqrt{3}+3\sqrt{48}$$

Combine 4\sqrt{3} and -6\sqrt{3} to get -2\sqrt{3}.

$$-2\sqrt{3}+3\sqrt{48}$$

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}.

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

Multiply 3 and 4 to get 12.

$$-2\sqrt{3}+12\sqrt{3}$$

Combine -2\sqrt{3} and 12\sqrt{3} to get 10\sqrt{3}.

$$10\sqrt{3}$$