Solve 9sqrt{3}(sqrt{6}+sqrt{14})

$$9 \sqrt { 3 } ( \sqrt { 6 } + \sqrt { 14 } )$$

Evaluate

$$9\left(\sqrt{42}+3\sqrt{2}\right)\approx 96.51043247$$

Solution Steps

Use the distributive property to multiply 9\sqrt{3} by \sqrt{6}+\sqrt{14}.

$$9\sqrt{3}\sqrt{6}+9\sqrt{3}\sqrt{14}$$

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}.

$$9\sqrt{3}\sqrt{3}\sqrt{2}+9\sqrt{3}\sqrt{14}$$

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

$$9\times 3\sqrt{2}+9\sqrt{3}\sqrt{14}$$

Multiply 9 and 3 to get 27.

$$27\sqrt{2}+9\sqrt{3}\sqrt{14}$$

To multiply \sqrt{3} and \sqrt{14}, multiply the numbers under the square root.

$$27\sqrt{2}+9\sqrt{42}$$