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$$9x^5+7x^2-9=0$$

How do we evaluate the roots of the given polynomial? We're asked to find its real positive zero.


What I tried doing:

Let $$f(x)=9x^5+7x^2-9$$ Using Descartes' rule of signs, I deduced that the given polynomial can have one positive real zero at the most. Also- $$f(-x)=-9x^5+7x^2-9$$ From this expression we can deduce that the given polynomial has two negative real zeros at the most. From here it is obvious that the polynomial has at least two complex roots.


This is not much progress, and I'm not sure what else I could do. According to Wolfram Alpha this polynomial has one real positive zero and four complex zeros. Any hints or explanation would be appreciated, thanks! Also, I'm not sure whether this is an irreducible polynomial or not, so I will not be including that tag.


EDIT:


I found the maxima and minima points of $f(x)$: $$f'(x)=45x^4+14x=0$$ $$x(45x^3+14)=0$$ $$x=0$$ or $$x=- \sqrt[3]{\frac {14}{45}}$$ This gives me the extremities of the curve, and I now have a rough approximation of the curve.

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    $\begingroup$ The polynomial is irreducible, yes. Descartes rule shows that we have at most one real root. Using the mean value theorem we see that there is exactly one (positive) real root. $\endgroup$ – Dietrich Burde May 4 at 18:09
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    $\begingroup$ You can use the Newton-Raphson method to get an approximativ solution $\endgroup$ – Dr. Sonnhard Graubner May 4 at 18:17
  • $\begingroup$ @Dr.SonnhardGraubner is this any other method? I've never heard of this method. Also I should mention that I'm (almost) a freshman at college. Any sources that explain this method would be appreciated! (I'll try to look it up myself anyway). $\endgroup$ – ExtremeRaider May 4 at 18:26
  • $\begingroup$ See here math.ubc.ca/~anstee/math104/104newtonmethod.pdf $\endgroup$ – Dr. Sonnhard Graubner May 4 at 18:33
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    $\begingroup$ @NikilKumar: you have to know that this is a dead end. You will run into terrible computation (the system will be nasty) and hit a wall. Because quintic equations were proved to have no general solution by radicals. $\endgroup$ – Yves Daoust May 5 at 7:07
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You are going to need to use numerical approximations. Newton-Raphson is good, there is also the Golden-Section search and the Bisection algorithm.

If you were wondering what the answer is, using mathematicas NSolve function will use a combination of Bisection, Newton-Raphson to approximate the values.

NSolve[$9x^5 + 7 x^2 - 9 == 0, x$]

Which produces the set

{
    {x -> -0.856287 - 0.435738 I}, 
    {x -> -0.856287 + 0.435738 I}, 
    {x -> 0.432007 - 1.04404 I}, 
    {x -> 0.432007 + 1.04404 I}, 
    {x -> 0.84856}
}

Resulting in your $1$ real solution of $f(0.84856) \sim 0$

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Observe that $f(0)=-9$ and $f(1)=7$. So by the IVT the polynomial has at least one real root. Now note that $f^\prime(x) = 45x^4+14x$. Creating a sign chart and sketching the graph, you observe that it crosses the $x$-axis exactly once. Therefore, you can conclude there is exactly one real root, and it is between $0$ and $1$.

Furthermore, applying the rational root theorem, the possible roots of $f$ are: $\pm 1, \pm 3,\pm 9, \pm \frac13, \pm \frac19$. Plugging each of these in, we find that none of them are roots of $f$. So the real-valued root of $f$ cannot be rational.

The Newton-Raphson method is just a specific application of techniques you learned in calculus. Since you know the real root is between $0$ and $1$, and this graph meets the sufficient conditions of convergence of this method, you can find the root for this graph. Furthermore, this method converges very rapidly to the root, so you can find a very precise approximation of this root within a few iterations.

You may be wondering: is there a way we can express the solution in the form of radicals, similar to the quadratic formula? The answer is no. According to the Impossibility Theorem, in general, there is no way to express the solutions of polynomials with degree five or higher in terms of radicals. So techniques like this are the best way to determine the roots of these equations.

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  • $\begingroup$ I believe that to use the Rational Root Theorem, you must start with a polynomial with integral coefficients. $\endgroup$ – Lubin May 4 at 18:53
  • $\begingroup$ @Lubin You are right, I got lazy. Fixed. $\endgroup$ – Hossmeister May 4 at 18:56
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    $\begingroup$ @Hossmeister I was not aware of the Impossibility Theorem at all. Thank you! $\endgroup$ – ExtremeRaider May 4 at 19:09
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$$f(0)=-9\text{ and }f(1)=7$$ so that the positive root lies somewhere in between. We can use the starting value $\frac12$ for Newton's iterations,

$$x\leftarrow x-\frac{9x^5+7x^2-9}{45x^4+14x}.$$

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    $\begingroup$ I believe the OP is looking for an exact solution. $\endgroup$ – Ovi May 4 at 18:45
  • $\begingroup$ @Ovi: this is a quintic equation, we don't expect a closed-form. $\endgroup$ – Yves Daoust May 4 at 18:50
  • $\begingroup$ Yes, but it appears this problem was an assignent from school, and it does not seem like the OP is studying approximations. So I expected there may be a way to cleverly add and subtract a term, which will allow you to factor. $\endgroup$ – Ovi May 5 at 2:17

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