Solve 2sqrt{27}*sqrt{32}:sqrt{48}

$$2 \sqrt { 27 } \times \sqrt { 32 } : \sqrt { 48 }$$

Evaluate

$$6\sqrt{2}\approx 8.485281374$$

Solution Steps

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

$$\frac{2\times 3\sqrt{3}\sqrt{32}}{\sqrt{48}}$$

Multiply 2 and 3 to get 6.

$$\frac{6\sqrt{3}\sqrt{32}}{\sqrt{48}}$$

Factor 32=4^{2}\times 2. Rewrite the square root of the product \sqrt{4^{2}\times 2} as the product of square roots \sqrt{4^{2}}\sqrt{2}. Take the square root of 4^{2}.

$$\frac{6\sqrt{3}\times 4\sqrt{2}}{\sqrt{48}}$$

Multiply 6 and 4 to get 24.

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

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

$$\frac{24\sqrt{6}}{\sqrt{48}}$$

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}.

$$\frac{24\sqrt{6}}{4\sqrt{3}}$$

Cancel out 4 in both numerator and denominator.

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

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

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

The square of \sqrt{3} is 3.

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

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}.

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

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

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

Cancel out 3 and 3.

$$6\sqrt{2}$$