Solve sqrt{20}-sqrt{600}+sqrt{125}-sqrt{54}=

$$\sqrt { 20 } - \sqrt { 600 } + \sqrt { 125 } - \sqrt { 54 } =$$

Evaluate

$$7\sqrt{5}-13\sqrt{6}\approx -16.190890814$$

Solution Steps

Factor 20=2^{2}\times 5. Rewrite the square root of the product \sqrt{2^{2}\times 5} as the product of square roots \sqrt{2^{2}}\sqrt{5}. Take the square root of 2^{2}.

$$2\sqrt{5}-\sqrt{600}+\sqrt{125}-\sqrt{54}$$

Factor 600=10^{2}\times 6. Rewrite the square root of the product \sqrt{10^{2}\times 6} as the product of square roots \sqrt{10^{2}}\sqrt{6}. Take the square root of 10^{2}.

$$2\sqrt{5}-10\sqrt{6}+\sqrt{125}-\sqrt{54}$$

Factor 125=5^{2}\times 5. Rewrite the square root of the product \sqrt{5^{2}\times 5} as the product of square roots \sqrt{5^{2}}\sqrt{5}. Take the square root of 5^{2}.

$$2\sqrt{5}-10\sqrt{6}+5\sqrt{5}-\sqrt{54}$$

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

$$7\sqrt{5}-10\sqrt{6}-\sqrt{54}$$

Factor 54=3^{2}\times 6. Rewrite the square root of the product \sqrt{3^{2}\times 6} as the product of square roots \sqrt{3^{2}}\sqrt{6}. Take the square root of 3^{2}.

$$7\sqrt{5}-10\sqrt{6}-3\sqrt{6}$$

Combine -10\sqrt{6} and -3\sqrt{6} to get -13\sqrt{6}.

$$7\sqrt{5}-13\sqrt{6}$$