Solve frac{5sqrt{18}+3sqrt{32}}{5sqrt{27}-3sqrt{12}}

$$\frac { 5 \sqrt { 18 } + 3 \sqrt { 32 } } { 5 \sqrt { 27 } - 3 \sqrt { 12 } }$$

Evaluate

$$\sqrt{6}\approx 2.449489743$$

Solution Steps

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}.

$$\frac{5\times 3\sqrt{2}+3\sqrt{32}}{5\sqrt{27}-3\sqrt{12}}$$

Multiply 5 and 3 to get 15.

$$\frac{15\sqrt{2}+3\sqrt{32}}{5\sqrt{27}-3\sqrt{12}}$$

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}.

$$\frac{15\sqrt{2}+3\times 4\sqrt{2}}{5\sqrt{27}-3\sqrt{12}}$$

Multiply 3 and 4 to get 12.

$$\frac{15\sqrt{2}+12\sqrt{2}}{5\sqrt{27}-3\sqrt{12}}$$

Combine 15\sqrt{2} and 12\sqrt{2} to get 27\sqrt{2}.

$$\frac{27\sqrt{2}}{5\sqrt{27}-3\sqrt{12}}$$

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}.

$$\frac{27\sqrt{2}}{5\times 3\sqrt{3}-3\sqrt{12}}$$

Multiply 5 and 3 to get 15.

$$\frac{27\sqrt{2}}{15\sqrt{3}-3\sqrt{12}}$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\frac{27\sqrt{2}}{15\sqrt{3}-3\times 2\sqrt{3}}$$

Multiply -3 and 2 to get -6.

$$\frac{27\sqrt{2}}{15\sqrt{3}-6\sqrt{3}}$$

Combine 15\sqrt{3} and -6\sqrt{3} to get 9\sqrt{3}.

$$\frac{27\sqrt{2}}{9\sqrt{3}}$$

Cancel out 9 in both numerator and denominator.

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

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

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

The square of \sqrt{3} is 3.

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

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

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

Cancel out 3 and 3.

$$\sqrt{6}$$