MHB What are the roots of a rational equation with given conditions?

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Find all irrational numbers $k$ such that $k^3-17k$ and $k^2+4k$ are both rational numbers.
 
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$k^2 + 4k = x$ be rational, following the question. Multiplying by $k$, we get $k^3 + 4k^2 = xk$. Substituting $k^2 = x - 4k$ in, one gets $k^3 + 4(x - 4k) = xk$. Thus, $k^3 - 16k = xk - 4x$ which implies $k^3 - 17k = xk - 4x - k = (x - 1)k - 4x$. But this thingy $k^3 - 17k$ is rational as per the question, which implies $(x - 1)k - 4x$ is rational. But $k$ is irrational and $x - 1$ and $4x$ are both rational ($x$ being rational). That'd imply $(x - 1)k - 4x$ is irrational, a contradiction if and only if $x - 1$ is nonzero. Otherwise, $x = 1$ and hence the desired irrationals are the roots of $k^2 + 4k = 1$.
 
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[sp]Let $k^2+4k = r$, rational. The quadratic equation $k^2+4k -r=0$ has solutions $k = -2 \pm\sqrt{4+r}$. Then $k^2 = 8+r \mp\sqrt{4+r}$ and $$k^3 - 17k = k(k^2-17) = (-2 \pm\sqrt{4+r})(-9+r \mp\sqrt{4+r}) = -2 -6r \pm(r-1)\sqrt{4+r}.$$ For that to be rational, we must have $r=1$ (because $\sqrt{4+r}$ must be irrational). That gives the solutions $k = -2\pm\sqrt5.$[/sp]
 
$k^2+4k$ is rational so $(k^2+ 4k + 4)$ or $(k+2)^2$

so $k = -2 \pm \sqrt{q}$ where q is rational but not square of rational

take $k = \sqrt{q}-2$ and
so we get
$k^3-17k$
= $(\sqrt{q} - 2)^3- 17(\sqrt{q}- 2)$
now take the irrational part to be zero to get
$q\sqrt{q} + 12\sqrt{q} - 17\sqrt{q}=0$ or q = 5
so one solution $\sqrt(5) - 2$
taking -ve q we get $-2-\sqrt{5}$
the 2 solutions are $\sqrt(5) - 2$ and $-2-\sqrt{5}$
 
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mathbalarka said:
$k^2 + 4k = x$ be rational, following the question. Multiplying by $k$, we get $k^3 + 4k^2 = xk$. Substituting $k^2 = x - 4k$ in, one gets $k^3 + 4(x - 4k) = xk$. Thus, $k^3 - 16k = xk - 4x$ which implies $k^3 - 17k = xk - 4x - k = (x - 1)k - 4x$. But this thingy $k^3 - 17k$ is rational as per the question, which implies $(x - 1)k - 4x$ is rational. But $k$ is irrational and $x - 1$ and $4x$ are both rational ($x$ being rational). That'd imply $(x - 1)k - 4x$ is irrational, a contradiction if and only if $x - 1$ is nonzero. Otherwise, $x = 1$ and hence the desired irrationals are the roots of $k^2 + 4k = 1$.

Hey mathbalarka, thanks for participating. Even though your conclusion is wrong, if you want to re-check your approach, just know that you're always welcome to re-submit of your most satisfied solution!(Yes)

Opalg said:
[sp]Let $k^2+4k = r$, rational. The quadratic equation $k^2+4k -r=0$ has solutions $k = -2 \pm\sqrt{4+r}$. Then $k^2 = 8+r \mp\sqrt{4+r}$ and $$k^3 - 17k = k(k^2-17) = (-2 \pm\sqrt{4+r})(-9+r \mp\sqrt{4+r}) = -2 -6r \pm(r-1)\sqrt{4+r}.$$ For that to be rational, we must have $r=1$ (because $\sqrt{4+r}$ must be irrational). That gives the solutions $k = -2\pm\sqrt5.$[/sp]

kaliprasad said:
$k^2+4k$ is rational so $(k^2+ 4k + 4)$ or $(k+2)^2$

so $k = -2 \pm \sqrt{q}$ where q is rational but not square of rational

take $k = \sqrt{q}-2$ and
so we get
$k^3-17k$
= $(\sqrt{q} - 2)^3- 17(\sqrt{q}- 2)$
now take the irrational part to be zero to get
$q\sqrt{q} + 12\sqrt{q} - 17\sqrt{q}=0$ or q = 5
so one solution $\sqrt(5) - 2$
taking -ve q we get $-2-\sqrt{5}$
the 2 solutions are $\sqrt(5) - 2$ and $-2-\sqrt{5}$

Thanks to both of you to participating in this challenge of mine and the answer is of course correct! Good job, Opalg! Bravo, kali!
 
Even though your conclusion is wrong, if you want to re-check your approach, just know that you're always welcome to re-submit of your most satisfied solution!

Ahem. I don't think my conclusion is wrong. The roots of $k^2 + 4k - 1 = 0$ are precisely $-2 \pm \sqrt{5}$, which satisfied your original conditions.
 
mathbalarka said:
Ahem. I don't think my conclusion is wrong. The roots of $k^2 + 4k - 1 = 0$ are precisely $-2 \pm \sqrt{5}$, which satisfied your original conditions.

Ah, I am sorry mathbalarka! I didn't realize you have edited your post, I responded to your solution based on what I read on your solution that sent to me through the email notification and I apologize.
 
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