Solve (3)(sqrt{24}-sqrt{12})*sqrt{1/3}

$$( 3 ) ( \sqrt { 24 } - \sqrt { 12 } ) \times \sqrt { \frac { 1 } { 3 } }$$

Evaluate

$$6\left(\sqrt{2}-1\right)\approx 2.485281374$$

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}.

$$3\left(2\sqrt{6}-\sqrt{12}\right)\sqrt{\frac{1}{3}}$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$3\left(2\sqrt{6}-2\sqrt{3}\right)\sqrt{\frac{1}{3}}$$

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

$$3\left(2\sqrt{6}-2\sqrt{3}\right)\times \frac{\sqrt{1}}{\sqrt{3}}$$

Calculate the square root of 1 and get 1.

$$3\left(2\sqrt{6}-2\sqrt{3}\right)\times \frac{1}{\sqrt{3}}$$

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

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

The square of \sqrt{3} is 3.

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

Cancel out 3 and 3.

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

Use the distributive property to multiply \sqrt{3} by 2\sqrt{6}-2\sqrt{3}.

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

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

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

Multiply \sqrt{3} and \sqrt{3} to get 3.

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

Multiply 2 and 3 to get 6.

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

The square of \sqrt{3} is 3.

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

Multiply -2 and 3 to get -6.

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