Solve (sqrt{18}-sqrt{12}+sqrt{2})*2sqrt{6}

$$( \sqrt { 18 } - \sqrt { 12 } + \sqrt { 2 } ) \times 2 \sqrt { 6 }$$

Evaluate

$$16\sqrt{3}-12\sqrt{2}\approx 10.742250173$$

Solution Steps

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}.

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

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}.

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

Combine 3\sqrt{2} and \sqrt{2} to get 4\sqrt{2}.

$$2\left(4\sqrt{2}-2\sqrt{3}\right)\sqrt{6}$$

Use the distributive property to multiply 2 by 4\sqrt{2}-2\sqrt{3}.

$$\left(8\sqrt{2}-4\sqrt{3}\right)\sqrt{6}$$

Use the distributive property to multiply 8\sqrt{2}-4\sqrt{3} by \sqrt{6}.

$$8\sqrt{2}\sqrt{6}-4\sqrt{3}\sqrt{6}$$

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

$$8\sqrt{2}\sqrt{2}\sqrt{3}-4\sqrt{3}\sqrt{6}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$8\times 2\sqrt{3}-4\sqrt{3}\sqrt{6}$$

Multiply 8 and 2 to get 16.

$$16\sqrt{3}-4\sqrt{3}\sqrt{6}$$

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}.

$$16\sqrt{3}-4\sqrt{3}\sqrt{3}\sqrt{2}$$

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

$$16\sqrt{3}-4\times 3\sqrt{2}$$

Multiply -4 and 3 to get -12.

$$16\sqrt{3}-12\sqrt{2}$$