Solve frac{sqrt{3}-sqrt{2}}{2sqrt{3}+3sqrt{2}}=

$$\frac { \sqrt { 3 } - \sqrt { 2 } } { 2 \sqrt { 3 } + 3 \sqrt { 2 } } =$$

Evaluate

$$\frac{5\sqrt{6}}{6}-2\approx 0.041241452$$

Solution Steps

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

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

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

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}-3\sqrt{2}\right)}{12-\left(3\sqrt{2}\right)^{2}}$$

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

$$\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}-3\sqrt{2}\right)}{12-3^{2}\left(\sqrt{2}\right)^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}-3\sqrt{2}\right)}{12-9\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}-3\sqrt{2}\right)}{12-9\times 2}$$

Multiply 9 and 2 to get 18.

$$\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(2\sqrt{3}-3\sqrt{2}\right)}{12-18}$$

Subtract 18 from 12 to get -6.

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 2 and 3 to get 6.

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

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

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

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

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

Combine -3\sqrt{6} and -2\sqrt{6} to get -5\sqrt{6}.

$$\frac{6-5\sqrt{6}+3\left(\sqrt{2}\right)^{2}}{-6}$$

The square of \sqrt{2} is 2.

$$\frac{6-5\sqrt{6}+3\times 2}{-6}$$

Multiply 3 and 2 to get 6.

$$\frac{6-5\sqrt{6}+6}{-6}$$

Add 6 and 6 to get 12.

$$\frac{12-5\sqrt{6}}{-6}$$

Multiply both numerator and denominator by -1.

$$\frac{-12+5\sqrt{6}}{6}$$