Solve {l}{sqrt{48}-7sqrt{12}}{+4sqrt{27}}

$$\left. \begin{array} { l } { \sqrt { 48 } - 7 \sqrt { 12 } } \\ { + 4 \sqrt { 27 } } \end{array} \right.$$

Evaluate

$$2\sqrt{3}\approx 3.464101615$$

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}.

$$4\sqrt{3}-7\sqrt{12}+4\sqrt{27}$$

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}.

$$4\sqrt{3}-7\times 2\sqrt{3}+4\sqrt{27}$$

Multiply -7 and 2 to get -14.

$$4\sqrt{3}-14\sqrt{3}+4\sqrt{27}$$

Combine 4\sqrt{3} and -14\sqrt{3} to get -10\sqrt{3}.

$$-10\sqrt{3}+4\sqrt{27}$$

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}.

$$-10\sqrt{3}+4\times 3\sqrt{3}$$

Multiply 4 and 3 to get 12.

$$-10\sqrt{3}+12\sqrt{3}$$

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

$$2\sqrt{3}$$