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if: $$L=\frac{\sqrt{5}+3}{3\sqrt{5}+k}$$ be a Rational number

then $k=?$

my try :

$$\frac{\sqrt{5}+3}{3\sqrt{5}+k}=\frac{(3+\sqrt{5})(k-3\sqrt{5})}{k^2-45}$$

then :?

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    Multiply top and group by $\sqrt{5}$ – 2017-01-24
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    $L=\frac13 \frac{3\sqrt{5}+9}{3\sqrt{5}+k}$, so ....? – 2017-01-24
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    Unless you specifically want to limit $k$ to be rational (or perhaps an integer), then I would say it's hard to pin down exactly what it can be. – 2017-01-24
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    @Andreas. then ?? – 2017-01-24
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    That's how I understood it: $k$ be integer. – 2017-01-24

4 Answers 4

1

Hint: write it as $(3L-1)\sqrt{5}=3 - k\,L\,$. If $\,3L-1 \ne 0\,$ then that would imply $\sqrt{5} = \frac{3 - k\,L}{3L-1} \in \mathbb{Q}\,$, but $\sqrt{5}$ is known to be irrational. Therefore $3L-1=0$ which gives $L=\frac{1}{3}\,$, then $k=\frac{3}{L}=9\,$.

[ EDIT ] The above assumes that the problem is looking for a rational $k \in \mathbb{Q}$.

If $k$ is not restricted to $\mathbb{Q}$, then for any $\forall L \in \mathbb{Q} \setminus \{0\}$ there exists a $k \in \mathbb{R}$ that satisfies the given equality, which can be calculated as: $\;k = \cfrac{\sqrt{5}+3}{L}- 3 \sqrt{5}=\cfrac{(1-3L)\sqrt{5} + 3}{L}\,$.

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    if $k \in$ Irrational number then ? – 2017-01-24
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    @Almot1960 I edited my answer to cover the case where $k$ is not restricted to $\mathbb{Q}\,$. – 2017-01-24
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    so : there are infinitely many solutions.? – 2017-01-24
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    @Almot1960 If $k$ is not restricted to $\mathbb{Q}$ then yes, there are infinitely many solutions. In fact, for each non-zero rational $L$ you can find a corresponding real $k\,$. – 2017-01-24
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Hint $ $ If $\,L\,$ is rational so too is $\,\dfrac{1}L\, =\, 3 + \dfrac{k-9}{\sqrt{5}+3}$

2

hint...Let the expression be $\frac pq$ where $p,q \in \mathbb{Z}$

Then equate rational and irrational parts to find $k\in\mathbb{Q}$. So from $$q\sqrt{5}+3q=3p\sqrt{5}+kp$$, we get $$q=3p$$ and $$3q=kp$$

Thus $$k=9$$

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    Please write complete . thank you very much . – 2017-01-24
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    Please see my edited answer – 2017-01-24
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    Is a unique k?? – 2017-01-24
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    No, there are infinitely many solutions. You assume $k$ is rational without comment. Without that you cannot equate the terms with and without $\sqrt 5$ – 2017-01-24
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    @RossMillikan.if : $k \in\mathbb{Q'}$ what ? – 2017-01-24
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    Then you can't separate the terms into the two equations as you did. See Bill Dubuque's answer. We just need the last term on the right to be rational. Another $k$ that works is $12+\sqrt 5$, for example. In general $k=9+r(\sqrt 5+3)$ where $r$ is any rational. – 2017-01-24
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An obvious candidate for $k$ is $9$ since for $k=9$ we obtain $$L=\frac{\sqrt{5}+3}{3\sqrt{5}+9}=\frac13$$

But should it be unique?

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    $k \in\mathbb{Q'}$ what ? – 2017-01-24