Given that $y=14+z$ and $y=z^4$ find the value of $y$.

Substituting $y=z^4$ we get:

$z^4 = 14 + z$ $\Rightarrow$ $z^4 - z - 14 = 0 $

I do not know how to approach solving this polynomial. The solutions I am looking for are real positive values of $y$.

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    $\begingroup$ Welcome to Math Stack Exchange. $z=2$ is a solution $\endgroup$ – J. W. Tanner Nov 19 at 23:26
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    $\begingroup$ @J.W.Tanner Can you please post your work? $\endgroup$ – nocomment Nov 19 at 23:30
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    $\begingroup$ Seeing $z=2$ as a solution is just done by inspection. $\endgroup$ – Andrew Chin Nov 19 at 23:33

I could tell by eye that $z=2$ is a solution.

(Looking at divisors of $14$ would be a more formal approach to find that.)

Now $(z^4-z-14)/(z-2)=z^3+2z^2+4z+7$,

which is $7$ when $z=0$ and strictly increasing for $z>0$,

so there are no other real positive solutions.

Correction in response to comment:

I showed that there are no other real positive solutions for $z$, but the question asked for real positive solutions for $y$, and there is another (irrational) one: $z^3+2z^2+4z+7 $ is negative $(1)$ when $z=-2$ and positive $(4)$ when $z=-1$, so there is a solution $z_0$ between $-2$ and $-1$, and $y_0=14+z_0$ is positive. Since $z^3+2z^2+4z+7$ is strictly increasing, there are no other real zeroes of $z^4-z-14$.

  • $\begingroup$ Thanks for posting. There seems to be a small error at $z^4-14$ instead of $z^4-z-14$ but I get the point $\endgroup$ – nocomment Nov 19 at 23:40
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    $\begingroup$ Thank you for catching the typo.; I corrected $\endgroup$ – J. W. Tanner Nov 19 at 23:41
  • $\begingroup$ The question only asks for real positive solutions for $y$, not $z$. The minimal value of $z$ that would ensure that $y$ is real and positive (or zero) is $z=-14$. $\endgroup$ – Glen O Nov 20 at 8:26
  • $\begingroup$ @GlenO With that value of $z$ you cannot satisfy the two equations simultaneously, though: the first one forces $y=0$, but then the second one $y=z^4$ does not hold. $\endgroup$ – Federico Poloni Nov 20 at 8:46
  • $\begingroup$ @FedericoPoloni - obviously $z=-14$ isn't a solution, but it makes the point that requiring $y\geq0$ doesn't mean $z\geq0$. In fact, there's a solution to $f(z)=z^3+2z^2+4z+7=0$ somewhere on $-2\leq z\leq -1$, as $f(-2)=-1$ and $f(-1)=4$. And for such a $z$, you get $12\leq y\leq 13$, which is definitely both real and positive. $\endgroup$ – Glen O Nov 20 at 12:57

There is a formula for quartic equations, but it is very complicated and pretty useless. Have you tried substituting $z$ for some divisor of $14$?

It has no more rational roots. It has no more positive roots, either.

  • $\begingroup$ Thank you. I have been able to see why $z=2$, as people have pointed out above. $\endgroup$ – nocomment Nov 19 at 23:36

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