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

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

Evaluate

$$3\sqrt{22}+6\sqrt{6}\approx 28.768185736$$

Solution Steps

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

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

Consider \left(2\sqrt{3}-\sqrt{11}\right)\left(2\sqrt{3}+\sqrt{11}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

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

The square of \sqrt{11} is 11.

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

Subtract 11 from 12 to get 1.

$$\frac{3\sqrt{2}\left(2\sqrt{3}+\sqrt{11}\right)}{1}$$

Anything divided by one gives itself.

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

Use the distributive property to multiply 3\sqrt{2} by 2\sqrt{3}+\sqrt{11}.

$$6\sqrt{3}\sqrt{2}+3\sqrt{2}\sqrt{11}$$

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

$$6\sqrt{6}+3\sqrt{2}\sqrt{11}$$

To multiply \sqrt{2} and \sqrt{11}, multiply the numbers under the square root.

$$6\sqrt{6}+3\sqrt{22}$$