Solve sqrt{18/8*27/2}

$$\sqrt { \frac { 18 } { 8 } \times \frac { 27 } { 2 } }$$

Evaluate

$$\frac{9\sqrt{6}}{4}\approx 5.511351921$$

Solution Steps

Reduce the fraction \frac{18}{8} to lowest terms by extracting and canceling out 2.

$$\sqrt{\frac{9}{4}\times \frac{27}{2}}$$

Multiply \frac{9}{4} times \frac{27}{2} by multiplying numerator times numerator and denominator times denominator.

$$\sqrt{\frac{9\times 27}{4\times 2}}$$

Do the multiplications in the fraction \frac{9\times 27}{4\times 2}.

$$\sqrt{\frac{243}{8}}$$

Rewrite the square root of the division \sqrt{\frac{243}{8}} as the division of square roots \frac{\sqrt{243}}{\sqrt{8}}.

$$\frac{\sqrt{243}}{\sqrt{8}}$$

Factor 243=9^{2}\times 3. Rewrite the square root of the product \sqrt{9^{2}\times 3} as the product of square roots \sqrt{9^{2}}\sqrt{3}. Take the square root of 9^{2}.

$$\frac{9\sqrt{3}}{\sqrt{8}}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\frac{9\sqrt{3}}{2\sqrt{2}}$$

Rationalize the denominator of \frac{9\sqrt{3}}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{9\sqrt{3}\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

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

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

$$\frac{9\sqrt{6}}{2\times 2}$$

Multiply 2 and 2 to get 4.

$$\frac{9\sqrt{6}}{4}$$