Solve frac{sqrt{36}}{sqrt{24}}+frac{sqrt{20}}{sqrt{81}}

$$\frac { \sqrt { 36 } } { \sqrt { 24 } } + \frac { \sqrt { 20 } } { \sqrt { 81 } }$$

Evaluate

$$\frac{\sqrt{6}}{2}+\frac{2\sqrt{5}}{9}\approx 1.721648866$$

Short Solution Steps

Calculate the square root of 36 and get 6.

$$\frac{6}{\sqrt{24}}+\frac{\sqrt{20}}{\sqrt{81}}$$

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}.

$$\frac{6}{2\sqrt{6}}+\frac{\sqrt{20}}{\sqrt{81}}$$

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

$$\frac{6\sqrt{6}}{2\left(\sqrt{6}\right)^{2}}+\frac{\sqrt{20}}{\sqrt{81}}$$

The square of \sqrt{6} is 6.

$$\frac{6\sqrt{6}}{2\times 6}+\frac{\sqrt{20}}{\sqrt{81}}$$

Cancel out 2\times 3 in both numerator and denominator.

$$\frac{\sqrt{6}}{2}+\frac{\sqrt{20}}{\sqrt{81}}$$

Factor 20=2^{2}\times 5. Rewrite the square root of the product \sqrt{2^{2}\times 5} as the product of square roots \sqrt{2^{2}}\sqrt{5}. Take the square root of 2^{2}.

$$\frac{\sqrt{6}}{2}+\frac{2\sqrt{5}}{\sqrt{81}}$$

Calculate the square root of 81 and get 9.

$$\frac{\sqrt{6}}{2}+\frac{2\sqrt{5}}{9}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and 9 is 18. Multiply \frac{\sqrt{6}}{2} times \frac{9}{9}. Multiply \frac{2\sqrt{5}}{9} times \frac{2}{2}.

$$\frac{9\sqrt{6}}{18}+\frac{2\times 2\sqrt{5}}{18}$$

Since \frac{9\sqrt{6}}{18} and \frac{2\times 2\sqrt{5}}{18} have the same denominator, add them by adding their numerators.

$$\frac{9\sqrt{6}+2\times 2\sqrt{5}}{18}$$

Do the multiplications in 9\sqrt{6}+2\times 2\sqrt{5}.

$$\frac{9\sqrt{6}+4\sqrt{5}}{18}$$