Solve frac{11sqrt{3}-3sqrt{2}}{3sqrt{2}-2sqrt{3}}

$$\frac { 11 \sqrt { 3 } - 3 \sqrt { 2 } } { 3 \sqrt { 2 } - 2 \sqrt { 3 } }$$

Evaluate

$$\frac{9\sqrt{6}}{2}+8\approx 19.022703843$$

Solution Steps

Rationalize the denominator of \frac{11\sqrt{3}-3\sqrt{2}}{3\sqrt{2}-2\sqrt{3}} by multiplying numerator and denominator by 3\sqrt{2}+2\sqrt{3}.

$$\frac{\left(11\sqrt{3}-3\sqrt{2}\right)\left(3\sqrt{2}+2\sqrt{3}\right)}{\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)}$$

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}.

$$\frac{\left(11\sqrt{3}-3\sqrt{2}\right)\left(3\sqrt{2}+2\sqrt{3}\right)}{\left(3\sqrt{2}\right)^{2}-\left(-2\sqrt{3}\right)^{2}}$$

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

$$\frac{\left(11\sqrt{3}-3\sqrt{2}\right)\left(3\sqrt{2}+2\sqrt{3}\right)}{3^{2}\left(\sqrt{2}\right)^{2}-\left(-2\sqrt{3}\right)^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{\left(11\sqrt{3}-3\sqrt{2}\right)\left(3\sqrt{2}+2\sqrt{3}\right)}{9\left(\sqrt{2}\right)^{2}-\left(-2\sqrt{3}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\left(11\sqrt{3}-3\sqrt{2}\right)\left(3\sqrt{2}+2\sqrt{3}\right)}{9\times 2-\left(-2\sqrt{3}\right)^{2}}$$

Multiply 9 and 2 to get 18.

$$\frac{\left(11\sqrt{3}-3\sqrt{2}\right)\left(3\sqrt{2}+2\sqrt{3}\right)}{18-\left(-2\sqrt{3}\right)^{2}}$$

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

$$\frac{\left(11\sqrt{3}-3\sqrt{2}\right)\left(3\sqrt{2}+2\sqrt{3}\right)}{18-\left(-2\right)^{2}\left(\sqrt{3}\right)^{2}}$$

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

$$\frac{\left(11\sqrt{3}-3\sqrt{2}\right)\left(3\sqrt{2}+2\sqrt{3}\right)}{18-4\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(11\sqrt{3}-3\sqrt{2}\right)\left(3\sqrt{2}+2\sqrt{3}\right)}{18-4\times 3}$$

Multiply 4 and 3 to get 12.

$$\frac{\left(11\sqrt{3}-3\sqrt{2}\right)\left(3\sqrt{2}+2\sqrt{3}\right)}{18-12}$$

Subtract 12 from 18 to get 6.

$$\frac{\left(11\sqrt{3}-3\sqrt{2}\right)\left(3\sqrt{2}+2\sqrt{3}\right)}{6}$$

Apply the distributive property by multiplying each term of 11\sqrt{3}-3\sqrt{2} by each term of 3\sqrt{2}+2\sqrt{3}.

$$\frac{33\sqrt{3}\sqrt{2}+22\left(\sqrt{3}\right)^{2}-9\left(\sqrt{2}\right)^{2}-6\sqrt{3}\sqrt{2}}{6}$$

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

$$\frac{33\sqrt{6}+22\left(\sqrt{3}\right)^{2}-9\left(\sqrt{2}\right)^{2}-6\sqrt{3}\sqrt{2}}{6}$$

The square of \sqrt{3} is 3.

$$\frac{33\sqrt{6}+22\times 3-9\left(\sqrt{2}\right)^{2}-6\sqrt{3}\sqrt{2}}{6}$$

Multiply 22 and 3 to get 66.

$$\frac{33\sqrt{6}+66-9\left(\sqrt{2}\right)^{2}-6\sqrt{3}\sqrt{2}}{6}$$

The square of \sqrt{2} is 2.

$$\frac{33\sqrt{6}+66-9\times 2-6\sqrt{3}\sqrt{2}}{6}$$

Multiply -9 and 2 to get -18.

$$\frac{33\sqrt{6}+66-18-6\sqrt{3}\sqrt{2}}{6}$$

Subtract 18 from 66 to get 48.

$$\frac{33\sqrt{6}+48-6\sqrt{3}\sqrt{2}}{6}$$

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

$$\frac{33\sqrt{6}+48-6\sqrt{6}}{6}$$

Combine 33\sqrt{6} and -6\sqrt{6} to get 27\sqrt{6}.

$$\frac{27\sqrt{6}+48}{6}$$