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

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

Evaluate

$$-\sqrt{6}-2\approx -4.449489743$$

Solution Steps

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

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

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}.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

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

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

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

Calculate -3 to the power of 2 and get 9.

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

The square of \sqrt{2} is 2.

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

Multiply 9 and 2 to get 18.

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

Subtract 18 from 12 to get -6.

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

Divide 2\sqrt{3}\left(2\sqrt{3}+3\sqrt{2}\right) by -6 to get -\frac{1}{3}\sqrt{3}\left(2\sqrt{3}+3\sqrt{2}\right).

$$-\frac{1}{3}\sqrt{3}\left(2\sqrt{3}+3\sqrt{2}\right)$$

Use the distributive property to multiply -\frac{1}{3}\sqrt{3} by 2\sqrt{3}+3\sqrt{2}.

$$-\frac{1}{3}\sqrt{3}\times 2\sqrt{3}-\frac{1}{3}\sqrt{3}\times 3\sqrt{2}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$-\frac{1}{3}\times 3\times 2-\frac{1}{3}\sqrt{3}\times 3\sqrt{2}$$

Cancel out 3 and 3.

$$-2-\frac{1}{3}\sqrt{3}\times 3\sqrt{2}$$

Cancel out 3 and 3.

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

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

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