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

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

Evaluate

$$\sqrt{6}+3\sqrt{2}+6\sqrt{3}+6\approx 23.084435275$$

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}.

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

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

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

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}.

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

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

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

Multiply 3 and 2 to get 6.

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

The square of \sqrt{2} is 2.

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

Multiply 3 and 2 to get 6.

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

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}.

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

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

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

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

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