Solve sqrt{12}+sqrt{27}-sqrt{48}=

$$\sqrt { 12 } + \sqrt { 27 } - \sqrt { 48 } =$$

Evaluate

$$\sqrt{3}\approx 1.732050808$$

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}.

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

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}.

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

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

$$5\sqrt{3}-\sqrt{48}$$

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}.

$$5\sqrt{3}-4\sqrt{3}$$

Combine 5\sqrt{3} and -4\sqrt{3} to get \sqrt{3}.

$$\sqrt{3}$$