Show that the largest root of $f$ is greater than $5n$ where $n(\ge 3)\in \mathbb N$.

Question

Given $f=x^3+9x^2+24x-40n^3+40xn^2+94n^2-12x^2n-62nx-74n+20$ has real roots, show that the largest root of $f$ is greater than $5n$ where $n(\ge 3)\in \mathbb N$.

I tried to do it by directly finding the roots in wolfram alpha for any $n\ge 3$, however I am getting the roots as complex which is against the hypothesis of the problem.

Also the roots obtained in Wolfram Alpha are very nasty or bad which is making my life my difficult.

Kindly help me out on how to find all the real roots of $f$ and show that they are $>5n$

Answer 1

Your polynomial equation is

$$\begin{equation}\begin{aligned} f(x)&=x^3+9x^2+24x-40n^3+40xn^2\\ & \; \; \; +94n^2-12x^2n-62nx-74n+20 \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

Note you have

$$\begin{equation}\begin{aligned} f(5n) & = (5n)^3 + 9(5n)^2 + 24(5n) - 40n^3 \\ & \; \; \; \; \; + 40(5n)n^2 + 94n^2 - 12(5n)^2n - 62n(5n) - 74n + 20 \\ & = 125n^3 + 225n^2 + 120n - 40n^3 + 200n^3 \\ & \; \; \; \; \; + 94n^2 - 300n^3 - 310n^2 - 74n + 20 \\ & = -15n^3 + 9n^2 + 46n + 20 \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

At $n = 3$, you thus get

$$\begin{equation}\begin{aligned} f(15) & = -15(3)^3 + 9(3)^2 + 46(3) + 20 \\ & = -166 \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

However, since the coefficient of the highest power of $f(x)$ is $1$ in $x^3$, this means that $\lim_{x \to \infty}f(x) = \infty$. Since $f(x)$ is continuous, there must be a root larger than $5n$ for when $n = 3$.

To confirm this is also true for all $n \gt 3$, one way is take the derivative of \eqref{eq2A}, as shown below

$$\frac{df(5n)}{dn} = -45n^2 + 18n + 46 \tag{4}\label{eq4A}$$

Using the quadratic formula to get the roots gives

$$\begin{equation}\begin{aligned} n & = \frac{-18 \pm \sqrt{18^2 - 4(-45)(46)}}{2(-45)} \\ & = \frac{3 \mp \sqrt{3^2 + 3(46)}}{15} \\ & = \frac{3 \mp \sqrt{3(3 + 46)}}{15} \\ & = \frac{3 \mp 7\sqrt{3}}{15} \\ & \approx -0.61, 1.01 \end{aligned}\end{equation}\tag{5}\label{eq5A}$$

The quadratic polynomial in \eqref{eq4A} being a concave-down parabola means its values are only positive with $n$ in the approximate range of $(-0.61,1.01)$, with it being negative everywhere else. Thus, for $n \ge 3$, the derivative is negative, so the value of \eqref{eq2A} would keep decreasing, confirming there's always a real root $\gt 5n$ for \eqref{eq1A}.

Answer 2

Hint: As the cubic is eventually positive, it is enough to show that $f(5n)<0$, i.e. $9n^2+46n+20 < 15n^3$. Can you show that, say using induction?

Answer 3

You have received good and simple answers and comments.

Thanks to the containment, let me give a more complex one. For sure, the formulae given by Wolfram Alpha or any other CAS are really messy. However, if you use the trigonometric method for solving the cubic, the results are not so bad.

For $$\left(-40 n^3+94 n^2-74 n+20\right)+\left(40 n^2-62 n+24\right) x+(9-12 n) x^2+x^3=0 $$ the roots are given by $$x_k=4n-3+\frac{2 \sqrt{8 n^2-10 n+3}}{\sqrt{3}}\times $$ $$\cos \left(\frac{2 \pi k}{3}-\frac{1}{3} \cos ^{-1}\left(3 \sqrt{3}\frac{ (n-1) \left(4 n^2-3 n+1\right)}{\left(8 n^2-10 n+3\right)^{3/2}}\right)\right)$$ with $k=0,1,2$. This is not so awful. The largest root (to prove) is $x_0$.

If you graph the roots as functions of $n$, you will notice that they are "almost" straight lines.

For the fun of it, compose Taylor series for large values of $n$; you will get $$x_0=\left(5+\sqrt{5}\right) n-\frac{75+11 \sqrt{5}}{20} -\frac{25-7 \sqrt{5}}{400\, n}-\frac{125+73 \sqrt{5}}{8000\, n^2}+O\left(\frac{1}{n^3}\right)$$

Consider the case where $n=3$; the exact solution is $$x_0=9+2 \sqrt{15} \cos \left(\frac{1}{3} \cos ^{-1}\left(\frac{56}{15 \sqrt{15}}\right)\right)\approx 16.7148$$ while the above truncated expansion gives $$\frac{808375+176747 \sqrt{5}}{72000}\approx 16.7166$$ This is not too bad.

Now, using the expansion $$\Delta=x_0-5n\sim n\sqrt{5} -\frac{75+11 \sqrt{5}}{20} $$ is positive as soon as $$n > \frac{11}{20}+\frac{3 \sqrt{5}}{4} \approx 2.22705$$

Thank you for the problem !

Edit

As I wrote in my answer, we need to prove that $x_0$ corresponds to the largest root of the cubic. The simplest way is to consider the series the series expansion for each of the roots. $$x_0=\left(5+\sqrt{5}\right) n-\frac{75+11 \sqrt{5}}{20} -\frac{25-7 \sqrt{5}}{400\, n}-\frac{125+73 \sqrt{5}}{8000\, n^2}+O\left(\frac{1}{n^3}\right)$$ $$x_1=\left(5-\sqrt{5}\right) n-\frac{75-11 \sqrt{5}}{20}-\frac{25+7 \sqrt{5}}{400 n}-\frac{125-73 \sqrt{5}}{8000\, n^2}+O\left(\frac{1}{n^3}\right)$$ $$x_2=2 n-\frac{3}{2}+\frac{1}{8 n}+\frac{1}{32 n^2}+O\left(\frac{1}{n^3}\right)$$ which clearly show the claim.

Using these truncated expressions, we properly find that $$x_0+x_1+x_2=12 n-9+O\left(\frac{1}{n^3}\right)$$ $$x_0\, x_1+x_0\, x_2+x_1\, x_2=40 n^2-62 n+24+O\left(\frac{1}{n^2}\right)$$ $$x_0\, x_1\, x_2=40 n^3-94 n^2+74 n-20+O\left(\frac{1}{n}\right)$$ which are exactly the coefficients of the cubic