Solve (sqrt{2}+sqrt{18})(sqrt{8}-sqrt{3})=

$$( \sqrt{ 2 } + \sqrt{ 18 } )( \sqrt{ 8 } - \sqrt{ 3 } )=$$

Evaluate

$$16-4\sqrt{6}\approx 6.202041029$$

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}.

$$\left(\sqrt{2}+3\sqrt{2}\right)\left(\sqrt{8}-\sqrt{3}\right)$$

Combine \sqrt{2} and 3\sqrt{2} to get 4\sqrt{2}.

$$4\sqrt{2}\left(\sqrt{8}-\sqrt{3}\right)$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$4\sqrt{2}\left(2\sqrt{2}-\sqrt{3}\right)$$

Use the distributive property to multiply 4\sqrt{2} by 2\sqrt{2}-\sqrt{3}.

$$8\left(\sqrt{2}\right)^{2}-4\sqrt{3}\sqrt{2}$$

The square of \sqrt{2} is 2.

$$8\times 2-4\sqrt{3}\sqrt{2}$$

Multiply 8 and 2 to get 16.

$$16-4\sqrt{3}\sqrt{2}$$

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

$$16-4\sqrt{6}$$