Solve 6/3sqrt{2-2sqrt{3}}

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

Evaluate

$$2\sqrt{3}+3\sqrt{2}\approx 7.706742302$$

Solution Steps

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

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

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}.

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

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

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

Calculate 3 to the power of 2 and get 9.

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

The square of \sqrt{2} is 2.

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

Multiply 9 and 2 to get 18.

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

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

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

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

Subtract 12 from 18 to get 6.

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

Cancel out 6 and 6.

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