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

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

Evaluate

$$4\sqrt{6}+6\sqrt{3}\approx 20.190263817$$

Solution Steps

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

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

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

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

Calculate 3 to the power of 2 and get 9.

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

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

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

Calculate -2 to the power of 2 and get 4.

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

The square of \sqrt{2} is 2.

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

Multiply 4 and 2 to get 8.

$$\frac{2\sqrt{3}\left(3+2\sqrt{2}\right)}{9-8}$$

Subtract 8 from 9 to get 1.

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

Anything divided by one gives itself.

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

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

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

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

$$6\sqrt{3}+4\sqrt{6}$$