Solve frac{sqrt{8}-2sqrt{12}+sqrt{32}}{sqrt{18}+sqrt{12}-sqrt{27}}

$$\frac { \sqrt { 8 } - 2 \sqrt { 12 } + \sqrt { 32 } } { \sqrt { 18 } + \sqrt { 12 } - \sqrt { 27 } }$$

Evaluate

$$\frac{8-2\sqrt{6}}{5}\approx 0.620204103$$

Solution Steps

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}.

$$\frac{2\sqrt{2}-2\sqrt{12}+\sqrt{32}}{\sqrt{18}+\sqrt{12}-\sqrt{27}}$$

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}.

$$\frac{2\sqrt{2}-2\times 2\sqrt{3}+\sqrt{32}}{\sqrt{18}+\sqrt{12}-\sqrt{27}}$$

Multiply -2 and 2 to get -4.

$$\frac{2\sqrt{2}-4\sqrt{3}+\sqrt{32}}{\sqrt{18}+\sqrt{12}-\sqrt{27}}$$

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}.

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

Combine 2\sqrt{2} and 4\sqrt{2} to get 6\sqrt{2}.

$$\frac{6\sqrt{2}-4\sqrt{3}}{\sqrt{18}+\sqrt{12}-\sqrt{27}}$$

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}.

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

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}.

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

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}.

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

Combine 2\sqrt{3} and -3\sqrt{3} to get -\sqrt{3}.

$$\frac{6\sqrt{2}-4\sqrt{3}}{3\sqrt{2}-\sqrt{3}}$$

Rationalize the denominator of \frac{6\sqrt{2}-4\sqrt{3}}{3\sqrt{2}-\sqrt{3}} by multiplying numerator and denominator by 3\sqrt{2}+\sqrt{3}.

$$\frac{\left(6\sqrt{2}-4\sqrt{3}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{\left(3\sqrt{2}-\sqrt{3}\right)\left(3\sqrt{2}+\sqrt{3}\right)}$$

Consider \left(3\sqrt{2}-\sqrt{3}\right)\left(3\sqrt{2}+\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(6\sqrt{2}-4\sqrt{3}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{\left(3\sqrt{2}\right)^{2}-\left(\sqrt{3}\right)^{2}}$$

Expand \left(3\sqrt{2}\right)^{2}.

$$\frac{\left(6\sqrt{2}-4\sqrt{3}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{3^{2}\left(\sqrt{2}\right)^{2}-\left(\sqrt{3}\right)^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(6\sqrt{2}-4\sqrt{3}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{9\left(\sqrt{2}\right)^{2}-\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\left(6\sqrt{2}-4\sqrt{3}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{9\times 2-\left(\sqrt{3}\right)^{2}}$$

Multiply 9 and 2 to get 18.

$$\frac{\left(6\sqrt{2}-4\sqrt{3}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{18-\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(6\sqrt{2}-4\sqrt{3}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{18-3}$$

Subtract 3 from 18 to get 15.

$$\frac{\left(6\sqrt{2}-4\sqrt{3}\right)\left(3\sqrt{2}+\sqrt{3}\right)}{15}$$

Apply the distributive property by multiplying each term of 6\sqrt{2}-4\sqrt{3} by each term of 3\sqrt{2}+\sqrt{3}.

$$\frac{18\left(\sqrt{2}\right)^{2}+6\sqrt{2}\sqrt{3}-12\sqrt{3}\sqrt{2}-4\left(\sqrt{3}\right)^{2}}{15}$$

The square of \sqrt{2} is 2.

$$\frac{18\times 2+6\sqrt{2}\sqrt{3}-12\sqrt{3}\sqrt{2}-4\left(\sqrt{3}\right)^{2}}{15}$$

Multiply 18 and 2 to get 36.

$$\frac{36+6\sqrt{2}\sqrt{3}-12\sqrt{3}\sqrt{2}-4\left(\sqrt{3}\right)^{2}}{15}$$

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

$$\frac{36+6\sqrt{6}-12\sqrt{3}\sqrt{2}-4\left(\sqrt{3}\right)^{2}}{15}$$

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

$$\frac{36+6\sqrt{6}-12\sqrt{6}-4\left(\sqrt{3}\right)^{2}}{15}$$

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

$$\frac{36-6\sqrt{6}-4\left(\sqrt{3}\right)^{2}}{15}$$

The square of \sqrt{3} is 3.

$$\frac{36-6\sqrt{6}-4\times 3}{15}$$

Multiply -4 and 3 to get -12.

$$\frac{36-6\sqrt{6}-12}{15}$$

Subtract 12 from 36 to get 24.

$$\frac{24-6\sqrt{6}}{15}$$