Solve sqrt{6}-sqrt{24}+sqrt{2}sqrt{3}

$$\sqrt{ 6 } - \sqrt{ 24 } + \sqrt{ 2 } \sqrt{ 3 }$$

Evaluate

$$0$$

Solution Steps

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}.

$$\sqrt{6}-2\sqrt{6}+\sqrt{2}\sqrt{3}$$

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

$$-\sqrt{6}+\sqrt{2}\sqrt{3}$$

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

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

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

$$0$$

Factor

$$0$$