Solve 10/3sqrt{2-sqrt{3}}

$$\frac { 10 } { 3 \sqrt { 2 } - \sqrt { 3 } }$$

Evaluate

$$\frac{2\sqrt{3}\left(\sqrt{6}+1\right)}{3}\approx 3.983127663$$

Solution Steps

Rationalize the denominator of \frac{10}{3\sqrt{2}-\sqrt{3}} by multiplying numerator and denominator by 3\sqrt{2}+\sqrt{3}.

$$\frac{10\left(3\sqrt{2}+\sqrt{3}\right)}{\left(3\sqrt{2}-\sqrt{3}\right)\left(3\sqrt{2}+\sqrt{3}\right)}$$

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

$$\frac{10\left(3\sqrt{2}+\sqrt{3}\right)}{\left(3\sqrt{2}\right)^{2}-\left(\sqrt{3}\right)^{2}}$$

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

$$\frac{10\left(3\sqrt{2}+\sqrt{3}\right)}{3^{2}\left(\sqrt{2}\right)^{2}-\left(\sqrt{3}\right)^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{10\left(3\sqrt{2}+\sqrt{3}\right)}{9\left(\sqrt{2}\right)^{2}-\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{10\left(3\sqrt{2}+\sqrt{3}\right)}{9\times 2-\left(\sqrt{3}\right)^{2}}$$

Multiply 9 and 2 to get 18.

$$\frac{10\left(3\sqrt{2}+\sqrt{3}\right)}{18-\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{10\left(3\sqrt{2}+\sqrt{3}\right)}{18-3}$$

Subtract 3 from 18 to get 15.

$$\frac{10\left(3\sqrt{2}+\sqrt{3}\right)}{15}$$

Divide 10\left(3\sqrt{2}+\sqrt{3}\right) by 15 to get \frac{2}{3}\left(3\sqrt{2}+\sqrt{3}\right).

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

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

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

Cancel out 3 and 3.

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