Solve (sqrt{18}-sqrt{6})*(sqrt{3}+1)

$$( \sqrt { 18 } - \sqrt { 6 } ) \times ( \sqrt { 3 } + 1 )$$

Evaluate

$$2\sqrt{6}\approx 4.898979486$$

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}.

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

Apply the distributive property by multiplying each term of 3\sqrt{2}-\sqrt{6} by each term of \sqrt{3}+1.

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

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

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

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}.

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

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

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

Combine 3\sqrt{2} and -3\sqrt{2} to get 0.

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

Combine 3\sqrt{6} and -\sqrt{6} to get 2\sqrt{6}.

$$2\sqrt{6}$$