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

[anemone]

Find all irrational numbers $k$ such that $k^3-17k$ and $k^2+4k$ are both rational numbers.

 

[mathbalarka]

Spoiler

$k^2 + 4k = x$ 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 $k^2 + 4k = 1$.

 

[Opalg]

[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]

Spoiler

$k^2+4k$ 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 $-2-\sqrt{5}$

 

[anemone]

Spoiler

$k^2 + 4k = x$ 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)

[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]

Spoiler

$k^2+4k$ 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!

 

[mathbalarka]

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.

 

[anemone]

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.