Solve sqrt{288}*sqrt{16}+frac{sqrt{24}}{3sqrt{3}}

$$\sqrt { 288 } \cdot \sqrt { 16 } + \frac { \sqrt { 24 } } { 3 \sqrt { 3 } }$$

Evaluate

$$\frac{146\sqrt{2}}{3}\approx 68.825060035$$

Solution Steps

Factor 288=16\times 18. Rewrite the square root of the product \sqrt{16\times 18} as the product of square roots \sqrt{16}\sqrt{18}.

$$\sqrt{16}\sqrt{18}\sqrt{16}+\frac{\sqrt{24}}{3\sqrt{3}}$$

Multiply \sqrt{16} and \sqrt{16} to get 16.

$$16\sqrt{18}+\frac{\sqrt{24}}{3\sqrt{3}}$$

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}.

$$16\times 3\sqrt{2}+\frac{\sqrt{24}}{3\sqrt{3}}$$

Multiply 16 and 3 to get 48.

$$48\sqrt{2}+\frac{\sqrt{24}}{3\sqrt{3}}$$

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}.

$$48\sqrt{2}+\frac{2\sqrt{6}}{3\sqrt{3}}$$

Rationalize the denominator of \frac{2\sqrt{6}}{3\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$48\sqrt{2}+\frac{2\sqrt{6}\sqrt{3}}{3\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$48\sqrt{2}+\frac{2\sqrt{6}\sqrt{3}}{3\times 3}$$

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}.

$$48\sqrt{2}+\frac{2\sqrt{3}\sqrt{2}\sqrt{3}}{3\times 3}$$

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

$$48\sqrt{2}+\frac{2\times 3\sqrt{2}}{3\times 3}$$

Multiply 3 and 3 to get 9.

$$48\sqrt{2}+\frac{2\times 3\sqrt{2}}{9}$$

Multiply 2 and 3 to get 6.

$$48\sqrt{2}+\frac{6\sqrt{2}}{9}$$

Divide 6\sqrt{2} by 9 to get \frac{2}{3}\sqrt{2}.

$$48\sqrt{2}+\frac{2}{3}\sqrt{2}$$

Combine 48\sqrt{2} and \frac{2}{3}\sqrt{2} to get \frac{146}{3}\sqrt{2}.

$$\frac{146}{3}\sqrt{2}$$