Solve sqrt{27}+sqrt{48}-sqrt{12}+2sqrt{3}

$$\sqrt { 27 } + \sqrt { 48 } - \sqrt { 12 } + 2 \sqrt { 3 }$$

Evaluate

$$7\sqrt{3}\approx 12.124355653$$

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}.

$$3\sqrt{3}+\sqrt{48}-\sqrt{12}+2\sqrt{3}$$

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}.

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

Combine 3\sqrt{3} and 4\sqrt{3} to get 7\sqrt{3}.

$$7\sqrt{3}-\sqrt{12}+2\sqrt{3}$$

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}.

$$7\sqrt{3}-2\sqrt{3}+2\sqrt{3}$$

Combine 7\sqrt{3} and -2\sqrt{3} to get 5\sqrt{3}.

$$5\sqrt{3}+2\sqrt{3}$$

Combine 5\sqrt{3} and 2\sqrt{3} to get 7\sqrt{3}.

$$7\sqrt{3}$$