Solve sqrt{243}+sqrt{27}-sqrt{75}=

$$\sqrt { 243 } + \sqrt { 27 } - \sqrt { 75 } =$$

Evaluate

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

Solution Steps

Factor 243=9^{2}\times 3. Rewrite the square root of the product \sqrt{9^{2}\times 3} as the product of square roots \sqrt{9^{2}}\sqrt{3}. Take the square root of 9^{2}.

$$9\sqrt{3}+\sqrt{27}-\sqrt{75}$$

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}.

$$9\sqrt{3}+3\sqrt{3}-\sqrt{75}$$

Combine 9\sqrt{3} and 3\sqrt{3} to get 12\sqrt{3}.

$$12\sqrt{3}-\sqrt{75}$$

Factor 75=5^{2}\times 3. Rewrite the square root of the product \sqrt{5^{2}\times 3} as the product of square roots \sqrt{5^{2}}\sqrt{3}. Take the square root of 5^{2}.

$$12\sqrt{3}-5\sqrt{3}$$

Combine 12\sqrt{3} and -5\sqrt{3} to get 7\sqrt{3}.

$$7\sqrt{3}$$