Problem Statement:-

Find all the values of $a$ for which the equation $$x^4+(a-1)x^3+x^2+(a-1)x+1=0$$ possess at least two negative roots.

I know that there has been a post regarding this same problem here, but I have a different issue with the problem than that posted in the OP in the above post.

So, I don't have an issue of finding a solution and coincidentally I had the same approach as the one posted by RicardoCruz.

But still if you have a solution than you are more than welcome to post one, well come on who doesn't want some good reputation change ;P.

My Attempt at a solution:-

For those who wanna skip to my main issue may start to read from the next quote.

We can see very clearly that $x=0$ isn't a root of the equation $$x^4 +(a-1)x^3 +x^2 +(a-1)x+1=0$$ hence it is safe to divide the equation throughout by $x^2$ for arriving at the condition which the problem wants.

So, we get $$\left(x^2+\dfrac{1}{x^2}+2\right)+(a-1)(x+\dfrac{1}{x})-1=0\\ \implies \left(x+\dfrac{1}{x}\right)^2+(a-1)(x+\dfrac{1}{x})-1=0$$.

Now, let $z=x+\dfrac{1}{x}$ then the equation becomes $$z^2+(a-1)z-1=0\\ \implies z=\dfrac{-(a-1)\pm\sqrt{\left(a-1\right)^2+4}}{2}$$

Now, its necessary to note that $\sqrt{\left(a-1\right)^2+4}\gt(a-1)$, hence for the roots to be negative $$\left(x+\dfrac{1}{x}\right)=\dfrac{-(a-1)-\sqrt{\left(a-1\right)^2+4}}{2}$$

As, we know that $\text{A.M. $\ge$ G.M.}$ for non negative real numbers, and as we are dealing with negative roots of the equation, so we will be applying $\text{A.M. $\ge$ G.M.}$ to $-x,-\dfrac{1}{x}$, we get $$\dfrac{-\left(x+\dfrac{1}{x}\right)}{2}\ge1\implies x+\dfrac{1}{x}\le-2\\ \implies \left(x+\dfrac{1}{x}\right)=\dfrac{-(a-1)-\sqrt{\left(a-1\right)^2+4}}{2}\le -2$$

But as the roots are supposed to be distinct hence, we have to exclude the case of equality, because it occurs when $x=\dfrac{1}{x}=-1$, which results in $x+\dfrac{1}{x}=-2\implies (x-1)^2=0$, hence $x$ has a repeated root. So, we get

$$\dfrac{-(a-1)-\sqrt{\left(a-1\right)^2+4}}{2}\lt -2\implies \sqrt{\left(a-1\right)^2+4}\gt 5-a$$

Now, this is where I am having problem, that's right solving the "inequality". I always had thought of this before but never tried to test it before, so bear with it and follow me on to a trip to solve inequalities in a weird way.

I did a case study for the inequality, which is as follows:-


If $(5-a)\gt 0\implies a\lt 5$, then there is no problem in squaring the inequality as both sides are positive. $$\left(a-1\right)^2+4\gt a^2+25-10a\implies a\gt \dfrac{5}{2}$$ $$\therefore a\in\left(\dfrac{5}{2},5\right)$$


If $5-a\lt0$, then we might be faced with a problem cause, there might be a case where $|5-a|\gt\sqrt{\left(a-1\right)^2+4}$, which would make the inequality $\sqrt{\left(a-1\right)^2+4}\gt 5-a$ on squaring change the sign of inequality to be in accordance with the condition $|5-a|\gt\sqrt{\left(a-1\right)^2+4}$.

So, on solving the inequality we get,

$$\sqrt{\left(a-1\right)^2+4}\gt 5-a\implies (a-1)^2+4\lt(5-a)^2$$ Now, why did I change the sign of inequality was due to the fact that I was finding the interval where(as per the condition I stated before) $|5-a|\gt\sqrt{\left(a-1\right)^2+4}$. So, just like $5\gt-6$ but on squaring the inequality we get $25\lt36$. So that's what I did.

This results in $a\lt\dfrac{5}{2}\cup a\gt5\implies a\in\emptyset$

So, on combining both the cases we get $a\in\left(\dfrac{5}{2},5\right)$

But from RobertCruz's solution and the books answer it seems that $ a\in\left(\dfrac{5}{2},\infty\right)$ is the correct solution, so what am I doing wrong here.

  • 1
    $\begingroup$ case 2: if 5−a < 0 Then the inequality holds for all eligible $a$ since the LHS > 0 > RHS. The eligible $a$ are determined by $5-a \lt 0 \iff a \gt 5$. $\endgroup$ – dxiv Nov 21 '16 at 7:33
  • $\begingroup$ @dxiv- isn't that what I have done O_o $\endgroup$ – user350331 Nov 21 '16 at 7:38
  • $\begingroup$ That's not what you wrote. You proceeded to square the inequality in some odd way which then gave a $a \lt \frac{5}{2}$ out of nowhere. If you just combine the ranges from cases 1 & 2 then $a \in (\frac{5}{2},5) \cup (5, \infty) = (\frac{5}{2},\infty) \setminus \{5\}$. That you are missing point $\{5\}$ is because you took all inequalities to be strict, while they needn't be. $\endgroup$ – dxiv Nov 21 '16 at 7:40
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    $\begingroup$ @user350331: As dxiv comments, if $5-a\color{red}{\le}0$, then $\sqrt{(a-1)^2+4}$ is positive and $5-a$ is non-positive. So, the inequality $\sqrt{(a-1)^2+4}\gt 5-a$ always holds. So, you don't need to try to square the both sides. Hence, the answer is $5/2\lt a\lt 5$ or $a\ge 5$. $\endgroup$ – mathlove Nov 21 '16 at 8:17
  • 1
    $\begingroup$ @user350331: It happens. Try to solve $\sqrt{a^2+a+1}\gt 2a$ both in the correct way and in the wrong way, and compare "the solutions". $\endgroup$ – mathlove Nov 21 '16 at 8:40

Just posting an alternative solution, as the comments already helped you to debug your solution:

For quartic polynomials, there is a closed-form analytical solution for the zeroes. However horrible that formula may seem, you don't need it here. You only need the discriminant.

A polynomial $ax^4+bx^3+cx^2+dx+e=0$ has two real solutions, when the discriminant is negative:

$$\Delta=256 a^3 e^3 - 192 a^2 b d e^2 - 128 a^2 c^2 e^2 + 144 a^2 c d^2 e$$ $$ - 27 a^2 d^4 + 144 a b^2 c e^2 - 6 a b^2 d^2 e - 80 a b c^2 d e + 18 a b c d^3 $$ $$+ 16 a c^4 e - 4 a c^3 d^2 - 27 b^4 e^2 + 18 b^3 c d e - 4 b^3 d^3 - 4 b^2 c^3 e + b^2 c^2 d^2<0$$

In your case, the polynomial can be written also as $$\underbrace{x^4+x^2+1}_{>0} + (a-1)\underbrace{(x+x^3)}_{>0 \text{ if } x>0}=0$$ for which any potential roots are negative if $a-1>0$ and positive otherwise.

Therefore, having checking for $a-1>0$ and negative discriminant ensures two negative real roots. You get a third degree polynomial in $u=(a-1)^2$.

$$144+8u-23u^2-4u^3<0$$ It turns out this factors: $$(4u-9)(4+u)^2>0$$ $$|u|>9/4$$ $$a-1>3/2$$

Also, you can quickly prove that your polynomial cannot have more than two real roots. Either by checking other tests on the link above (checking the value for $P$ and $D$) or by writing $$x=-\frac{1}{1-a}\frac{1+x^2+x^4}{1+x^2}=-\frac{1}{a-1}\left(1+\frac{x^4}{1+x^2}\right)$$ RHS has no inflection points, so a straight line can intersect it twice at most (definition of convexity).


From the place where you have $x+1/x\; (=z)$ in terms of $a,$ write $x+1/x=r$ where $$r=(-(a-1)-\sqrt {(a-1)^2+4})/2 <0.$$ So $x^2-rx+1=0,$ giving $$x=(r\pm \sqrt {r^2-4}\;)/2<0.$$ This yields two negative roots iff $r^2-4>0,$ that is, iff $|r|>2$ because $r^2-4<0$ yields no real negative $x$, and $r^2-4=0$ yields exactly one negative root. In terms of $a,$ this is $$2<|(-(a-1)-\sqrt {(a-1)^2+4}\;)/2|=|(\;a-1+\sqrt {(a-1)^2+4}\;)/2| .$$

(Because we must have $a>1,$ else we have $ x^4+(a-1)x^3+x^2+(a-1)x+1>0$ whenever $x<0$.)

For $a>1$ this is equivalent to $a-1>3/2,$ that is, $a>5/2.$

Another method would be: Let $f(x)=x^4+(a-1)x^3+x^2+(a-1)x+1$ with $a>1.$ Then $$(f(x)<0\land x<0)\iff$$ $$ a-1=-\frac {x^4+x^2+1}{-x^3-x}=$$ $$=\frac {(x+1/x)^2x^2-x^2}{-x^2(x+1/x)}=$$ $$=\frac {u^2-1}{u}$$ where $u=|x|+|1/x|=-x-1/x.$

The AGM inequality implies that $|x|+|1/x|\geq 2.$ Now show that $\frac {u^2-1}{u}\geq \frac {3}{2}$ for $u\geq 2.$ And I leave the rest to you.

Remark. For $u\geq 2$ let $u=2+b.$ Then $\frac {u^2-1}{u}-\frac {3}{2}=\frac {8b^2+2b}{2b+4}.$


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