The real positive root of $9x^5+7x^2-9=0$

Question

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

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

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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.

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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.

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EDIT:

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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.

Answer 1

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 $f(0.84856) \sim 0$

Answer 2

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.

Answer 3

$$f(0)=-9\text{ and }f(1)=7$$ so that the positive root lies somewhere in between. We can use the starting value $\frac12$ for Newton's iterations,

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