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

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

Evaluate

$$2\sqrt{6}-5\approx -0.101020514$$

Solution Steps

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

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

Consider \left(3\sqrt{2}+2\sqrt{3}\right)\left(3\sqrt{2}-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}.

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

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

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

Calculate 3 to the power of 2 and get 9.

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

The square of \sqrt{2} is 2.

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

Multiply 9 and 2 to get 18.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

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

Subtract 12 from 18 to get 6.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Multiply -4 and 3 to get -12.

$$\frac{6\sqrt{6}-12-9\left(\sqrt{2}\right)^{2}+6\sqrt{3}\sqrt{2}}{6}$$

The square of \sqrt{2} is 2.

$$\frac{6\sqrt{6}-12-9\times 2+6\sqrt{3}\sqrt{2}}{6}$$

Multiply -9 and 2 to get -18.

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

Subtract 18 from -12 to get -30.

$$\frac{6\sqrt{6}-30+6\sqrt{3}\sqrt{2}}{6}$$

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

$$\frac{6\sqrt{6}-30+6\sqrt{6}}{6}$$

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

$$\frac{12\sqrt{6}-30}{6}$$

Divide each term of 12\sqrt{6}-30 by 6 to get 2\sqrt{6}-5.

$$2\sqrt{6}-5$$

Factor

$$2 \sqrt{6} - 5 = -0.101020514$$