Question

6+√27/4-√3 can be written in the form r +s√3/13

Answer 1

$r = \left(\dfrac{33}{13}\right)$

$s = \left(\dfrac{10}{13}\right)$

Step-by-step explanation:

Let's multiply and divide the given fraction by the conjugate of the denominator:

$\dfrac{6+\sqrt{27}}{4-\sqrt{3}}×\dfrac{4+\sqrt{3}}{4+\sqrt{3}}$

$\;\;\;\;= \dfrac{24 + 6\sqrt{3} + 4\sqrt{3} + \sqrt{27}\sqrt{3}}{13}$

$\;\;\;\;=\frac{1}{13}(24 + 10\sqrt{3} + \sqrt{81})$

$\;\;\;\;=\frac{1}{13}(33 + 10\sqrt{3})$

$\;\;\;\;=\left(\dfrac{33}{13}\right) + \left(\dfrac{10}{13}\right)\dfrac{\sqrt{3}}{13}$

We can see here that

$r = \left(\dfrac{33}{13}\right)$

$s = \left(\dfrac{10}{13}\right)$