Solve frac{3sqrt{32}-2sqrt{18}}{4sqrt{27}+2sqrt{3}}

$$\frac { 3 \sqrt { 32 } - 2 \sqrt { 18 } } { 4 \sqrt { 27 } + 2 \sqrt { 3 } }$$

Evaluate

$$\frac{\sqrt{6}}{7}\approx 0.349927106$$

Solution Steps

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}.

$$\frac{3\times 4\sqrt{2}-2\sqrt{18}}{4\sqrt{27}+2\sqrt{3}}$$

Multiply 3 and 4 to get 12.

$$\frac{12\sqrt{2}-2\sqrt{18}}{4\sqrt{27}+2\sqrt{3}}$$

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}.

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

Multiply -2 and 3 to get -6.

$$\frac{12\sqrt{2}-6\sqrt{2}}{4\sqrt{27}+2\sqrt{3}}$$

Combine 12\sqrt{2} and -6\sqrt{2} to get 6\sqrt{2}.

$$\frac{6\sqrt{2}}{4\sqrt{27}+2\sqrt{3}}$$

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}.

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

Multiply 4 and 3 to get 12.

$$\frac{6\sqrt{2}}{12\sqrt{3}+2\sqrt{3}}$$

Combine 12\sqrt{3} and 2\sqrt{3} to get 14\sqrt{3}.

$$\frac{6\sqrt{2}}{14\sqrt{3}}$$

Cancel out 2 in both numerator and denominator.

$$\frac{3\sqrt{2}}{7\sqrt{3}}$$

Rationalize the denominator of \frac{3\sqrt{2}}{7\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$\frac{3\sqrt{2}\sqrt{3}}{7\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{3\sqrt{2}\sqrt{3}}{7\times 3}$$

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

$$\frac{3\sqrt{6}}{7\times 3}$$

Multiply 7 and 3 to get 21.

$$\frac{3\sqrt{6}}{21}$$

Divide 3\sqrt{6} by 21 to get \frac{1}{7}\sqrt{6}.

$$\frac{1}{7}\sqrt{6}$$