Solve (2sqrt{3}-sqrt{18})(sqrt{6}-1)

$$( 2 \sqrt { 3 } - \sqrt { 18 } ) ( \sqrt { 6 } - 1 )$$

Evaluate

$$9\sqrt{2}-8\sqrt{3}\approx -1.128484399$$

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}.

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

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

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

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}.

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

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

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

Multiply 2 and 3 to get 6.

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

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}.

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

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

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

Multiply -3 and 2 to get -6.

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

Combine -2\sqrt{3} and -6\sqrt{3} to get -8\sqrt{3}.

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

Combine 6\sqrt{2} and 3\sqrt{2} to get 9\sqrt{2}.

$$9\sqrt{2}-8\sqrt{3}$$