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

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

Evaluate

$$-6\sqrt{3}-10\approx -20.392304845$$

Solution Steps

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

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

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}.

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

Square \sqrt{3}. Square 2.

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

Subtract 4 from 3 to get -1.

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

Anything divided by -1 gives its opposite.

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 2 and 3 to get 6.

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

Add 4 and 6 to get 10.

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

Combine 2\sqrt{3} and 4\sqrt{3} to get 6\sqrt{3}.

$$-\left(6\sqrt{3}+10\right)$$

To find the opposite of 6\sqrt{3}+10, find the opposite of each term.

$$-6\sqrt{3}-10$$

Factor

$$2 {(-3 \sqrt{3} - 5)} = -20.392304845$$