Solve (3sqrt{6}+sqrt{24})divsqrt{2}

$$( 3 \sqrt { 6 } + \sqrt { 24 } ) \div \sqrt { 2 }$$

Evaluate

$$5\sqrt{3}\approx 8.660254038$$

Solution Steps

Factor 24=2^{2}\times 6. Rewrite the square root of the product \sqrt{2^{2}\times 6} as the product of square roots \sqrt{2^{2}}\sqrt{6}. Take the square root of 2^{2}.

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

Combine 3\sqrt{6} and 2\sqrt{6} to get 5\sqrt{6}.

$$\frac{5\sqrt{6}}{\sqrt{2}}$$

Rationalize the denominator of \frac{5\sqrt{6}}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{5\sqrt{6}\sqrt{2}}{\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{5\sqrt{6}\sqrt{2}}{2}$$

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}.

$$\frac{5\sqrt{2}\sqrt{3}\sqrt{2}}{2}$$

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

$$\frac{5\times 2\sqrt{3}}{2}$$

Cancel out 2 and 2.

$$5\sqrt{3}$$