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

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

Evaluate

$$\frac{\sqrt{2}+\sqrt{6}-2\sqrt{3}-1}{5}\approx -0.120079662$$

Solution Steps

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

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

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

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

The square of \sqrt{2} is 2.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

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

Subtract 12 from 2 to get -10.

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

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

$$\frac{\left(\sqrt{2}\right)^{2}-2\sqrt{2}\sqrt{3}-2\sqrt{2}+4\sqrt{3}}{-10}$$

The square of \sqrt{2} is 2.

$$\frac{2-2\sqrt{2}\sqrt{3}-2\sqrt{2}+4\sqrt{3}}{-10}$$

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

$$\frac{2-2\sqrt{6}-2\sqrt{2}+4\sqrt{3}}{-10}$$