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

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

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Answer 1 · 2017-04-30T19:41:50

$a = 4; b = 1$ $a=4"; "b=1$

Explanation:

Assumption: The question should be:

$\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} = a + \sqrt{15} . b$ $(sqrt5+sqrt3)/(sqrt5-sqrt3)=a+sqrt(15)color(white)(.) b$

$Consider the left hand side (LHS) only$ $color(blue)("Consider the left hand side (LHS) only")$

$\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} \times 1$ $color(green)((sqrt5+sqrt3)/(sqrt5-sqrt3)color(red)(xx1) )$

$\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} \times \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}}$ $color(green)((sqrt5+sqrt3)/(sqrt5-sqrt3)color(red)(xx (sqrt5+sqrt3)/(sqrt5+sqrt3) )$

$\frac{(\sqrt{5} + \sqrt{3}) (\sqrt{5} + \sqrt{3})}{5 - 3}$ $((sqrt5+sqrt3)(sqrt5+sqrt3))/(5-3)$

$\frac{5 + 2 \sqrt{3 \times 5} + 3}{2}$ $(5+2sqrt(3xx5)+3)/2$

$\frac{8}{2} + \frac{2 \sqrt{15}}{2}$ $8/2+(2sqrt(15))/2$

$4 + \sqrt{15}$ $4+sqrt15$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Putting it all together$ $color(blue)("Putting it all together")$

$4 + \sqrt{15} = a + \sqrt{15} . . b$ $4+sqrt15=a+sqrt15color(white)(..)b$

Thus $a = 4; b = 1$ $a=4"; "b=1$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Additional note$ $color(blue)("Additional note")$

If the RHS $\to a + \sqrt{15 b . .}$ $->a+sqrt(15bcolor(white)(..))$

The $b$ $b$ is still 1

Answer 2 · 2017-04-30T19:45:42

$a = 4$ $a=4" "$ and $b = 1$ $" "b=1$

Explanation:

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

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

$\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} = \frac{{(\sqrt{5})}^{2} + 2 \sqrt{5} \sqrt{3} + {(\sqrt{3})}^{2}}{{(\sqrt{5})}^{2} - {(\sqrt{3})}^{2}}$ $color(white)((sqrt(5)+sqrt(3))/(sqrt(5)-sqrt(3))) = ((sqrt(5))^2+2sqrt(5)sqrt(3)+(sqrt(3))^2)/((sqrt(5))^2-(sqrt(3))^2)$

$\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} = \frac{5 + 2 \sqrt{15} + 3}{5 - 3}$ $color(white)((sqrt(5)+sqrt(3))/(sqrt(5)-sqrt(3))) = (5+2sqrt(15)+3)/(5-3)$

$\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} = \frac{8 + 2 \sqrt{15}}{2}$ $color(white)((sqrt(5)+sqrt(3))/(sqrt(5)-sqrt(3))) = (8+2sqrt(15))/2$

$\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} = 4 + \sqrt{15}$ $color(white)((sqrt(5)+sqrt(3))/(sqrt(5)-sqrt(3))) = 4+sqrt(15)$

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

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

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