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

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

Evaluate

$$\frac{7-2\sqrt{6}}{5}\approx 0.420204103$$

Solution Steps

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

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

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

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

Multiply 2\sqrt{3}-\sqrt{2} and 2\sqrt{3}-\sqrt{2} to get \left(2\sqrt{3}-\sqrt{2}\right)^{2}.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{3}-\sqrt{2}\right)^{2}.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$\frac{12-4\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}}{\left(2\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

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

$$\frac{12-4\sqrt{6}+\left(\sqrt{2}\right)^{2}}{\left(2\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{12-4\sqrt{6}+2}{\left(2\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

Add 12 and 2 to get 14.

$$\frac{14-4\sqrt{6}}{\left(2\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

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

$$\frac{14-4\sqrt{6}}{2^{2}\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{14-4\sqrt{6}}{4\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{14-4\sqrt{6}}{4\times 3-\left(\sqrt{2}\right)^{2}}$$

Multiply 4 and 3 to get 12.

$$\frac{14-4\sqrt{6}}{12-\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{14-4\sqrt{6}}{12-2}$$

Subtract 2 from 12 to get 10.

$$\frac{14-4\sqrt{6}}{10}$$

Factor

$$\frac{7 - 2 \sqrt{6}}{5} = 0.4202041028867288$$