Solve (2sqrt{18}-3sqrt{27})+sqrt{6}quad(11)(sqrt{3})

$$( 2 \sqrt { 18 } - 3 \sqrt { 27 } ) + \sqrt { 6 } \quad ( 11 ) ( \sqrt { 3 } )$$

Evaluate

$$39\sqrt{2}-9\sqrt{3}\approx 39.565871664$$

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}.

$$2\times 3\sqrt{2}-3\sqrt{27}+\sqrt{6}\times 11\sqrt{3}$$

Multiply 2 and 3 to get 6.

$$6\sqrt{2}-3\sqrt{27}+\sqrt{6}\times 11\sqrt{3}$$

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

$$6\sqrt{2}-3\times 3\sqrt{3}+\sqrt{6}\times 11\sqrt{3}$$

Multiply -3 and 3 to get -9.

$$6\sqrt{2}-9\sqrt{3}+\sqrt{6}\times 11\sqrt{3}$$

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}.

$$6\sqrt{2}-9\sqrt{3}+\sqrt{3}\sqrt{2}\times 11\sqrt{3}$$

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

$$6\sqrt{2}-9\sqrt{3}+3\times 11\sqrt{2}$$

Multiply 3 and 11 to get 33.

$$6\sqrt{2}-9\sqrt{3}+33\sqrt{2}$$

Combine 6\sqrt{2} and 33\sqrt{2} to get 39\sqrt{2}.

$$39\sqrt{2}-9\sqrt{3}$$