Solve (2sqrt{3}+sqrt{48})*3sqrt{2}div2

$$(2 \sqrt{ 3 } + \sqrt{ 48 } ) \times 3 \sqrt{ 2 } \div 2$$

Evaluate

$$9\sqrt{6}\approx 22.045407685$$

Solution Steps

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

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

Combine 2\sqrt{3} and 4\sqrt{3} to get 6\sqrt{3}.

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

Multiply 3 and 6 to get 18.

$$\frac{18\sqrt{3}\sqrt{2}}{2}$$

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

$$\frac{18\sqrt{6}}{2}$$

Divide 18\sqrt{6} by 2 to get 9\sqrt{6}.

$$9\sqrt{6}$$