If (sqrt(5)+sqrt(3))/(sqrt(5)-sqrt(3)) = a+sqrt(15)b for integers a and b then what are a and b ?

Apr 30, 2017

$a = 4 \text{; } b = 1$

Explanation:

Assumption: The question should be:

$\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} = a + \sqrt{15} \textcolor{w h i t e}{.} b$

$\textcolor{b l u e}{\text{Consider the left hand side (LHS) only}}$

$\textcolor{g r e e n}{\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} \textcolor{red}{\times 1}}$

color(green)((sqrt5+sqrt3)/(sqrt5-sqrt3)color(red)(xx (sqrt5+sqrt3)/(sqrt5+sqrt3) )

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

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

$\frac{8}{2} + \frac{2 \sqrt{15}}{2}$

$4 + \sqrt{15}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Putting it all together}}$

$4 + \sqrt{15} = a + \sqrt{15} \textcolor{w h i t e}{. .} b$

Thus $a = 4 \text{; } b = 1$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Additional note}}$

If the RHS $\to a + \sqrt{15 b \textcolor{w h i t e}{. .}}$

The $b$ is still 1

Apr 30, 2017

$a = 4 \text{ }$ and $\text{ } b = 1$

Explanation:

To rational the denominator $\sqrt{5} - \sqrt{3}$ we can multiply both numerator and denominator by the radical conjugate $\sqrt{5} + \sqrt{3}$...

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

$\textcolor{w h i t e}{\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}} = \frac{{\left(\sqrt{5}\right)}^{2} + 2 \sqrt{5} \sqrt{3} + {\left(\sqrt{3}\right)}^{2}}{{\left(\sqrt{5}\right)}^{2} - {\left(\sqrt{3}\right)}^{2}}$

$\textcolor{w h i t e}{\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}} = \frac{5 + 2 \sqrt{15} + 3}{5 - 3}$

$\textcolor{w h i t e}{\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}} = \frac{8 + 2 \sqrt{15}}{2}$

$\textcolor{w h i t e}{\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}} = 4 + \sqrt{15}$

So $a = 4$ and $b = 1$