How to find roots of a 4th order polynomial?

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 $$y$$.

• Welcome to Math Stack Exchange. $z=2$ is a solution – J. W. Tanner Nov 19 at 23:26
• @J.W.Tanner Can you please post your work? – nocomment Nov 19 at 23:30
• Seeing $z=2$ as a solution is just done by inspection. – 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 $$z^4-z-14$$.

• 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 – nocomment Nov 19 at 23:40
• Thank you for catching the typo.; I corrected – J. W. Tanner Nov 19 at 23:41
• 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 $z=-14$. – Glen O Nov 20 at 8:26
• @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. – Federico Poloni Nov 20 at 8:46
• @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. – 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.

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