Solve frac{6+sqrt{27}}{4-sqrt{3}}

$$\frac { 6 + \sqrt { 27 } } { 4 - \sqrt { 3 } }$$

Evaluate

$$\frac{18\sqrt{3}+33}{13}\approx 4.936685734$$

Solution Steps

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

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

Rationalize the denominator of \frac{6+3\sqrt{3}}{4-\sqrt{3}} by multiplying numerator and denominator by 4+\sqrt{3}.

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

Consider \left(4-\sqrt{3}\right)\left(4+\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}.

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

Square 4. Square \sqrt{3}.

$$\frac{\left(6+3\sqrt{3}\right)\left(4+\sqrt{3}\right)}{16-3}$$

Subtract 3 from 16 to get 13.

$$\frac{\left(6+3\sqrt{3}\right)\left(4+\sqrt{3}\right)}{13}$$

Apply the distributive property by multiplying each term of 6+3\sqrt{3} by each term of 4+\sqrt{3}.

$$\frac{24+6\sqrt{3}+12\sqrt{3}+3\left(\sqrt{3}\right)^{2}}{13}$$

Combine 6\sqrt{3} and 12\sqrt{3} to get 18\sqrt{3}.

$$\frac{24+18\sqrt{3}+3\left(\sqrt{3}\right)^{2}}{13}$$

The square of \sqrt{3} is 3.

$$\frac{24+18\sqrt{3}+3\times 3}{13}$$

Multiply 3 and 3 to get 9.

$$\frac{24+18\sqrt{3}+9}{13}$$

Add 24 and 9 to get 33.

$$\frac{33+18\sqrt{3}}{13}$$