Solve 3sqrt{2}+3sqrt{2}*sqrt{18}-2sqrt{18}

$$3 \sqrt { 2 } + 3 \sqrt { 2 } \cdot \sqrt { 18 } - 2 \sqrt { 18 }$$

Evaluate

$$18-3\sqrt{2}\approx 13.757359313$$

Solution Steps

Factor 18=2\times 9. Rewrite the square root of the product \sqrt{2\times 9} as the product of square roots \sqrt{2}\sqrt{9}.

$$3\sqrt{2}+3\sqrt{2}\sqrt{2}\sqrt{9}-2\sqrt{18}$$

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

$$3\sqrt{2}+3\times 2\sqrt{9}-2\sqrt{18}$$

Multiply 3 and 2 to get 6.

$$3\sqrt{2}+6\sqrt{9}-2\sqrt{18}$$

Calculate the square root of 9 and get 3.

$$3\sqrt{2}+6\times 3-2\sqrt{18}$$

Multiply 6 and 3 to get 18.

$$3\sqrt{2}+18-2\sqrt{18}$$

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}.

$$3\sqrt{2}+18-2\times 3\sqrt{2}$$

Multiply -2 and 3 to get -6.

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

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

$$-3\sqrt{2}+18$$