Solve (sqrt{600}+sqrt{0}-sqrt{24})*sqrt{6}

$$( \sqrt { 600 } + \sqrt { 0 } - \sqrt { 24 } ) \cdot \sqrt { 6 }$$

Evaluate

$$48$$

Solution Steps

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}.

$$\left(10\sqrt{6}+\sqrt{0}-\sqrt{24}\right)\sqrt{6}$$

Calculate the square root of 0 and get 0.

$$\left(10\sqrt{6}+0-\sqrt{24}\right)\sqrt{6}$$

Factor 24=2^{2}\times 6. Rewrite the square root of the product \sqrt{2^{2}\times 6} as the product of square roots \sqrt{2^{2}}\sqrt{6}. Take the square root of 2^{2}.

$$\left(10\sqrt{6}+0-2\sqrt{6}\right)\sqrt{6}$$

Combine 10\sqrt{6} and -2\sqrt{6} to get 8\sqrt{6}.

$$\left(8\sqrt{6}+0\right)\sqrt{6}$$

Anything plus zero gives itself.

$$8\sqrt{6}\sqrt{6}$$

Multiply \sqrt{6} and \sqrt{6} to get 6.

$$8\times 6$$

Multiply 8 and 6 to get 48.

$$48$$

Factor

$$2^{4}\times 3$$