Solve 6/sqrt{3}+frac{3+sqrt{3}}{sqrt{5}-2}=

$$\frac { 6 } { \sqrt { 3 } } + \frac { 3 + \sqrt { 3 } } { \sqrt { 5 } - 2 } =$$

Evaluate

$$\sqrt{15}+3\sqrt{5}+4\sqrt{3}+6\approx 23.509390509$$

Solution Steps

Rationalize the denominator of \frac{6}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

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

The square of \sqrt{3} is 3.

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

Divide 6\sqrt{3} by 3 to get 2\sqrt{3}.

$$2\sqrt{3}+\frac{3+\sqrt{3}}{\sqrt{5}-2}$$

Rationalize the denominator of \frac{3+\sqrt{3}}{\sqrt{5}-2} by multiplying numerator and denominator by \sqrt{5}+2.

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

Consider \left(\sqrt{5}-2\right)\left(\sqrt{5}+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}.

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

Square \sqrt{5}. Square 2.

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

Subtract 4 from 5 to get 1.

$$2\sqrt{3}+\frac{\left(3+\sqrt{3}\right)\left(\sqrt{5}+2\right)}{1}$$

Anything divided by one gives itself.

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

Apply the distributive property by multiplying each term of 3+\sqrt{3} by each term of \sqrt{5}+2.

$$2\sqrt{3}+3\sqrt{5}+6+\sqrt{3}\sqrt{5}+2\sqrt{3}$$

To multiply \sqrt{3} and \sqrt{5}, multiply the numbers under the square root.

$$2\sqrt{3}+3\sqrt{5}+6+\sqrt{15}+2\sqrt{3}$$

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

$$4\sqrt{3}+3\sqrt{5}+6+\sqrt{15}$$