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

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

Evaluate

$$2\sqrt{2}+2\sqrt{3}-\sqrt{6}-4\approx -0.156961003$$

Solution Steps

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

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

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

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

Square \sqrt{3}. Square 2.

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

Subtract 4 from 3 to get -1.

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

Anything divided by -1 gives its opposite.

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

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

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

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

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

To find the opposite of \sqrt{6}-2\sqrt{2}-2\sqrt{3}+4, find the opposite of each term.

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

The opposite of -2\sqrt{2} is 2\sqrt{2}.

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

The opposite of -2\sqrt{3} is 2\sqrt{3}.

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