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

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

Evaluate

$$\sqrt{6}-2\approx 0.449489743$$

Solution Steps

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

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

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

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

The square of \sqrt{3} is 3.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{2} is 2.

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

Multiply 4 and 2 to get 8.

$$\frac{\left(2\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-2\sqrt{2}\right)}{3-8}$$

Subtract 8 from 3 to get -5.

$$\frac{\left(2\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-2\sqrt{2}\right)}{-5}$$

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

$$\frac{2\left(\sqrt{3}\right)^{2}-4\sqrt{3}\sqrt{2}-\sqrt{2}\sqrt{3}+2\left(\sqrt{2}\right)^{2}}{-5}$$

The square of \sqrt{3} is 3.

$$\frac{2\times 3-4\sqrt{3}\sqrt{2}-\sqrt{2}\sqrt{3}+2\left(\sqrt{2}\right)^{2}}{-5}$$

Multiply 2 and 3 to get 6.

$$\frac{6-4\sqrt{3}\sqrt{2}-\sqrt{2}\sqrt{3}+2\left(\sqrt{2}\right)^{2}}{-5}$$

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

$$\frac{6-4\sqrt{6}-\sqrt{2}\sqrt{3}+2\left(\sqrt{2}\right)^{2}}{-5}$$

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

$$\frac{6-4\sqrt{6}-\sqrt{6}+2\left(\sqrt{2}\right)^{2}}{-5}$$

Combine -4\sqrt{6} and -\sqrt{6} to get -5\sqrt{6}.

$$\frac{6-5\sqrt{6}+2\left(\sqrt{2}\right)^{2}}{-5}$$

The square of \sqrt{2} is 2.

$$\frac{6-5\sqrt{6}+2\times 2}{-5}$$

Multiply 2 and 2 to get 4.

$$\frac{6-5\sqrt{6}+4}{-5}$$

Add 6 and 4 to get 10.

$$\frac{10-5\sqrt{6}}{-5}$$

Divide each term of 10-5\sqrt{6} by -5 to get -2+\sqrt{6}.

$$-2+\sqrt{6}$$

Factor

$$\sqrt{6} - 2 = 0.449489743$$