Solve sqrt{24}-sqrt{18}sqrt{1/3}

$$\sqrt{ 24 } - \sqrt{ 18 } \sqrt{ \frac{ 1 }{ 3 } }$$

Evaluate

$$\sqrt{6}\approx 2.449489743$$

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}.

$$2\sqrt{6}-\sqrt{18}\sqrt{\frac{1}{3}}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$2\sqrt{6}-3\sqrt{2}\sqrt{\frac{1}{3}}$$

Rewrite the square root of the division \sqrt{\frac{1}{3}} as the division of square roots \frac{\sqrt{1}}{\sqrt{3}}.

$$2\sqrt{6}-3\sqrt{2}\times \frac{\sqrt{1}}{\sqrt{3}}$$

Calculate the square root of 1 and get 1.

$$2\sqrt{6}-3\sqrt{2}\times \frac{1}{\sqrt{3}}$$

Rationalize the denominator of \frac{1}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$2\sqrt{6}-3\sqrt{2}\times \frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$2\sqrt{6}-3\sqrt{2}\times \frac{\sqrt{3}}{3}$$

Cancel out 3 and 3.

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

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

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

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

$$\sqrt{6}$$