Solve -3sqrt{48}+2sqrt{24}+2sqrt{6}+4sqrt{12}

$$- 3 \sqrt { 48 } + 2 \sqrt { 24 } + 2 \sqrt { 6 } + 4 \sqrt { 12 }$$

Evaluate

$$2\sqrt{3}\left(3\sqrt{2}-2\right)\approx 7.768735226$$

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}.

$$-3\times 4\sqrt{3}+2\sqrt{24}+2\sqrt{6}+4\sqrt{12}$$

Multiply -3 and 4 to get -12.

$$-12\sqrt{3}+2\sqrt{24}+2\sqrt{6}+4\sqrt{12}$$

Factor 24=2^{2}\times 6. Rewrite the square root of the product \sqrt{2^{2}\times 6} as the product of square roots \sqrt{2^{2}}\sqrt{6}. Take the square root of 2^{2}.

$$-12\sqrt{3}+2\times 2\sqrt{6}+2\sqrt{6}+4\sqrt{12}$$

Multiply 2 and 2 to get 4.

$$-12\sqrt{3}+4\sqrt{6}+2\sqrt{6}+4\sqrt{12}$$

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

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

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}.

$$-12\sqrt{3}+6\sqrt{6}+4\times 2\sqrt{3}$$

Multiply 4 and 2 to get 8.

$$-12\sqrt{3}+6\sqrt{6}+8\sqrt{3}$$

Combine -12\sqrt{3} and 8\sqrt{3} to get -4\sqrt{3}.

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