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$2x^{4}+5ax^{3}-2bx^{2}+1=0$ has no real root in $(5,2014)$
Find the conditions for $a$ and $b$

I am suspicious of even the existence of its solution and at a loss.

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What do you mean by "the existence of its solution" - the solution to this problem, or this polynomial? Remember that by the fundamental theorem of algebra, this polynomial does have a root in the complex numbers $\mathbb{C}$ (in fact, 4 roots when counting multiplicity). The issue is just whether any of those roots are real numbers, and whether they lie in that range. – Zev Chonoles May 16 '13 at 7:00
By continuity, the function $f(x)=2x^4+5ax^3-2bx^2+1$ has a real root in $(5, 2014)$ either if one of local max or local min inside this interval is zero ( think about x^2-100) or $f(x)$ has different signs (think about x^3-216). Now the problem is basically taking derivative, finding relative max and min, then setting up some inequalities. – Jiangwei Xue May 16 '13 at 7:07
@Headhunter -- I'd appreciate it if you could look at the source of the question and see if $a>0$ and $b>0$ are assumed in the question. Otherwise it seems peculiar that there be $+5ax^3$ and $-2bx^2$ terms in the equation, since allowing any signs on $a,b$ would make the extra minus unnecessary. I think one can show something in case $a,b>0$ and wouldn't like to answer without knowing whether $a,b>0$ is actually an assumption. – coffeemath May 18 '13 at 10:06
up vote 2 down vote accepted

This answer assumes $a,b>0$. (If it were otherwise, the presence of the minus sign in the $-2bx^2$ term seems unnecessary). If this is not a valid assumption, my result at least treats one of the four possibilities; note that if $a>0,b<0$ the equation has no positive roots at all, so really only two cases remain.

Anyway let $p(x)=2x^4+5ax^3-2bx^2+1$ and note that since $p(0)=1$ we may define $q(x)=p(x)/x^2$ and $q(x)$ will have the same zeroes as $p(x).$ $$q(x)=5ax-2b +2x^2+x^{-2},\\ q'(x)=5a+4x-2x^{-3}.$$ Since $a>0$ we see that $q'(x)>0$ for $x \in [5,2014].$ [in fact $q'>0$ on $((1/2)^{1/4},\infty)].$ This means that $q$ is increasing on $[5,2014]$ so that $q$ maps this interval bijectively (and continuously) to the interval $[q(5),q(2014)].$ Now the condition that there be no zero in the open interval $(5,2014)$ becomes equivalent to saying that either $q(5) \ge 0$ or $q(2014) \le 0.$ Luckily both expressions for $q(5)$ and for $q(2014)$ involve the same subtracted term $-2b$; in fact $$q(5)=25a-2b+1251/25,\ \ q(2014)=10070a-2b+k,$$ where $k$ is an ungainly fraction. We can then express the conditions on $a,b$ for no zero in $(5,2014)$ by saying that $b$ satisfies, in terms of $a$, one of the inequalities $$b \le (25/2)a+1251/50,$$ which corresponds to $q(5) \ge 0$, or else $$b \ge 5035a+k_2,$$ where $k_2 \approx 4.056\cdot 10^6$, which corresponds to $q(2014) \le 0.$

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The region(s) in the $ab$-plane where $(a,b)$ leads to $p_{(a,b)}(x)=2x^4+5ax^3-2bx^2+1$ having a root inside $(5,2013)$ is bordered by curves (possibly straight lines) where one of the following happens:

  • $p_{(a,b)}$ has a root at 5
  • $p_{(a,b)}$ has a root at 2013
  • $p_{(a,b)}$ has a double root inside $(5,2013)$
  • $p_{(a,b)}$ has a quadruple root inside $(5,2013)$

This is because if there is a root inside $(5,2013)$ and we perturb $(a,b)$ by $\epsilon$, these are the only ways that the root could suddenly disappear from the interval $(5,2013)$.

So I suggest mapping out the curves that define these boundaries and then examining the regions they contain. The first two boundary curves are quickly seen to be straight lines defined by setting $p(5)=0$ and $p(2013)=0$. A quadruple root is not possible since $p$'s linear term is $0$, and the only way for that to happen with a fourth degree polynomial that has a quadruple root is if the polynomial is $cx^4$.

So the difficult curve is the one that defines where $p_{(a,b)}$ has a double root inside $(5,2013)$. For $p$ to have a double root means $p'$ has that root with multiplicity one. $p'$ in this problem has been made rather easy to analyze: $p'(x)=x(8x^2+15ax-4b)$. Clearly the common root between $p$ and $p'$ would not be $0$, so a double root would be one of the numbers $\frac{-15a\pm\sqrt{225a^2+128b}}{16}$. Plugging this back into $p(x)=0$, we have a quartic relation in $a$ and $b$: $$2\left(\frac{-15a\pm\sqrt{225a^2+128b}}{16}\right)^4+5a\left(\frac{-15a\pm\sqrt{225a^2+128b}}{16}\right)^3-2b\left(\frac{-15a\pm\sqrt{225a^2+128b}}{16}\right)^2+1=0$$

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