Solve sqrt{108}-sqrt{108}+sqrt{12}=

$$\sqrt { 108 } - \sqrt { 108 } + \sqrt { 12 } =$$

Evaluate

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

Solution Steps

Factor 108=6^{2}\times 3. Rewrite the square root of the product \sqrt{6^{2}\times 3} as the product of square roots \sqrt{6^{2}}\sqrt{3}. Take the square root of 6^{2}.

$$6\sqrt{3}-\sqrt{108}+\sqrt{12}$$

Factor 108=6^{2}\times 3. Rewrite the square root of the product \sqrt{6^{2}\times 3} as the product of square roots \sqrt{6^{2}}\sqrt{3}. Take the square root of 6^{2}.

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

Combine 6\sqrt{3} and -6\sqrt{3} to get 0.

$$\sqrt{12}$$

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}.

$$2\sqrt{3}$$