Solve sqrt{6}(sqrt{3}+sqrt{2}-sqrt{18})

$$\sqrt { 6 } ( \sqrt { 3 } + \sqrt { 2 } - \sqrt { 18 } )$$

Evaluate

$$\sqrt{3}\left(\sqrt{6}-4\right)\approx -2.685562543$$

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}.

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

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

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

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

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

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}.

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

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

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

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}.

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

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

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

Multiply -2 and 2 to get -4.

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